$$ \DeclareMathOperator{\Span}{span} \DeclareMathOperator{\Det}{det} \DeclareMathOperator{\Tr}{tr} \DeclareMathOperator{\Adj}{adj} \DeclareMathOperator{\Dim}{dim} \DeclareMathOperator{\Rank}{rank} \DeclareMathOperator{\Ker}{ker} \DeclareMathOperator{\Img}{im} \DeclareMathOperator{\Diag}{diag} \DeclareMathOperator{\Proj}{Proj} $$

Linear Algebra

When working with data, we typically deal with many data points consisting of many dimensions. That is, each data point may have a several components; e.g. if people are your data points, they may be represented by their height, weight, and age, which constitutes three dimensions all together. These data points of many components are called vectors. These are contrasted with individual values, which are called scalars.

We deal with these data points - vectors - simultaneously in aggregates known as matrices.

Linear algebra provides the tools needed to manipulate vectors and matrices.


Vectors have magnitude and direction, e.g. 5 miles/hour going east. The magnitude can be thought of, in some sense, as the "length" of the vector (this isn't quite right however, as there are many concepts of "length" - see norms).

A vector
A vector

Formally, this example would be represented:

$$ \vec{v} = \begin{bmatrix} 5 \\ 0 \end{bmatrix} $$

since we are "moving" 5 on x-axis and 0 on the y-axis.

Note that often the arrow is dropped, i.e. the vector is notated as just $v$.

Real coordinate spaces

Vectors are plotted and manipulated in space. A two-dimensional vector, such as the previous example, may be represented in a two-dimensional space.

A vector with three components would be represented in a three-dimensional space, and so on for any arbitrary $n$ dimensions.

A real coordinate space (that is, a space consisting of real numbers) of $n$ dimensions is notated $\mathbb R^n$. Such a space encapsulates all possible vectors of that dimensionality, i.e. all possible vectors of the form $[v_1, v_2, \dots, v_n]$.

To denote a vector of $n$ dimensions, we write $x \in \mathbb R^n$.

For example: the notation for the two-dimensional real coordinate space is notated $\mathbb R^2$, which is all possible real-valued 2-tuples (i.e. all 2D vectors whose components are real numbers, e.g. $[1, 2], [-0.4, 21.4], \dots$). If we wanted to describe an arbitrary two-dimensional vector, we could do so with $\vec v \in \mathbb R^2$.

Column and row vectors

A vector $x \in \mathbb R^n$ typically denotes a column vector, i.e. with $n$ rows and 1 column.

A row vector $x^T \in \mathbb R^n$ has 1 row and $n$ columns. The notation $x^T$ is described below.

Transposing a vector

Transposing a vector means turning its rows into columns:

$$ \vec{a} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}, \\ \vec{a}^T = \begin{bmatrix} x_1 & x_2 & x_3 & x_4 \end{bmatrix} $$

So a column vector $x$ can be represented as a row vector with $x^T$.

Vectors operations

Vector addition

Vectors are added by adding the individual corresponding components:

$$ \begin{bmatrix} 6 \\ 2 \end{bmatrix} + \begin{bmatrix} -4 \\ 4 \end{bmatrix} = \begin{bmatrix} 6 + -4 \\ 2 + 4 \end{bmatrix} = \begin{bmatrix} 2 \\ 6 \end{bmatrix} $$

Multiplying a vector by a scalar

Example: The red vector is before multiplying a scalar, blue is after.
Example: The red vector is before multiplying a scalar, blue is after.

To multiply a vector with a scalar, you just multiply the individual components of the vector by the scalar:

$$ 3\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \times 2 \\ 3 \times 1 \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \end{bmatrix} $$

This changes the magnitude of the vector, but not the direction.

Vector dot products

The dot product (also called inner product) of two vectors $\vec{a}, \vec{b} \in \mathbb R^n$ (note that this implies they must be of the same dimension) is notated:

$$ \vec{a} \cdot \vec{b} $$

It is calculated:

$$ \vec{a} \cdot \vec{b} = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} \cdot \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} = a_1b_1 + a_2b_2 + \dots + a_nb_n = \sum^n_{i=1} a_i b_i $$

Which results in a scalar value.

Note that sometimes the dot operator is dropped, so a dot product may be notated as just $\vec{a} \vec{b}$.

Also note that the dot product of $x \cdot y$ is equivalent to the matrix multiplication $x^Ty$.

Properties of vector dot products:


The norm of a vector $x \in \mathbb R^n$, denoted $||x||$, is the "length" of the vector. That is, norms are a generalization of "distance" or "length".

There are many different norms, the most common of which is the Euclidean norm (also known as the $\ell_2$ norm), denoted $||x||_2$, computed:

$$ ||x||_2 = \sqrt{ \sum^n_{i=1} x^2_i } = \sqrt{x^Tx}$$

This is the "as-the-crow-flies" distance that we are all familiar with.

Generally, a norm is just any function $f : \mathbb R^n \to \mathbb R$ which satisfies the following properties:

  1. non-negativity: For all $x \in \mathbb R^n$, $f(x) \geq 0$
  2. definiteness: $f(x) = 0$ if and only if $x=0$
  3. homogeneity: For all $x \in \mathbb R^n, t \in \mathbb R, f(tx) = |t|f(x)$
  4. triangle inequality: For all $x,y \in \mathbb R^n, f(x+y) \leq f(x) + f(y)$

Another norm is the $\ell_1$ norm:

$$ ||x||_1 = \sum^n_{i=1} |x_i| $$

and the $\ell_{\infty}$ norm:

$$ ||x||_{\infty} = \max_i|x_i| $$

These three norms are part of the family of $\ell_p$ norms, which are parameterized by a real number $p \geq 1$ and defined as:

$$ ||x||_p = (\sum^n_{i=1} |x_i|^p)^{\frac{1}{p}} $$

There are also norms for [matrices], the most common of which is the Frobenius norm, analogous to the Euclidean ($\ell_2$) norm for vectors:

$$ ||A||_{\text{Fro}} = \sqrt{\sum_{n=1}^N \sum_{m=1}^M A_{n,m}^2} $$

Lengths and dot products

You may notice that the dot product of a vector with itself is the square of that vector's length:

$$ \vec{a} \cdot \vec{a} = a_1^2 + a_2^2 + \dots + a_n^2 = ||\vec{a}||^2 $$

So the length (Euclidean norm) of a vector can be written:

$$ ||\vec{a}|| = \sqrt{\vec{a} \cdot \vec{a}} $$

Unit vectors

Each dimension in a space has a unit vector, generally denoted with a hat, e.g. $\hat u$, which is a vector constrained to that dimension (that is, it has 0 magnitude in all other dimensions), with length 1, e.g. $||\hat u|| = 1$.

Unit vectors exists for all $\mathbb R^n$.

The unit vector is also called a normalized vector (which is not to be confused with a normal vector, which is something else entirely.)

The unit vector in the same direction as some vector
$\vec{v}$ is found by computing:

$$ \hat u = \frac{\vec{v}}{||\vec{v}||} $$

For instance, in $\mathbb R^2$ space, we would have two unit vectors:

$$ \hat{i} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} ,\; \hat{j} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$

In $\mathbb R^3$ space, we would have three unit vectors:

$$ \hat{i} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} ,\; \hat{j} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} ,\; \hat{k} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$

But you can have unit vectors in any direction. Say you have a vector:

$$ \vec{a} = \begin{bmatrix} 5 \\ -6 \end{bmatrix} $$

You can find a unit vector $\hat{u}$ in the direction of this vector like so:

$$ \hat{u} = \frac{\vec{a}}{||\vec{a}||} $$

so, with our example:

$$ \hat{u} = \frac{1}{||\vec{a}||}\vec{a} = \frac{1}{\sqrt{61}}\begin{bmatrix}5 \\ -6\end{bmatrix} = \begin{bmatrix} \frac{5}{\sqrt{61}} \\ \frac{-6}{\sqrt{61}} \end{bmatrix}$$

Angles between vectors

Say you have two non-zero vectors, $\vec{a}, \vec{b} \in \mathbb R^n$.

We often notate the angle between two vectors as $\theta$.

Angle between two vectors.
Angle between two vectors.

The law of cosine tells us, that for a triangle:

$$ C^2 = A^2 + B^2 - 2AB\cos\theta $$

Using this law, we can get the angle between our two vectors:

$$ ||\vec{a} - \vec{b}||^2 = ||\vec{b}||^2 + ||\vec{a}||^2 - 2||\vec{a}||||\vec{b}||\cos\theta $$

which simplifies to:

$$ \vec{a} \cdot \vec{b} = ||\vec{a}||||\vec{b}||\cos\theta $$

There are two special cases if the vectors are collinear, that is if $\vec{a} = c\vec{b}$:

Perpendicular vectors

With the above angle calculation, you can see that if $\vec{a}$ and $\vec{b}$ are non-zero, and their dot product is 0, that is, $\vec{a} \cdot \vec{b} = \vec{0}$, then they are perpendicular to each other.

Whenever a pair of vectors satisfies this condition $\vec{a} \cdot \vec{b} = \vec{0}$, it is said that the two vectors are orthogonal (i.e. perpendicular).

Note that because any vector times the zero vector equals the zero vector: $\vec{0} \cdot \vec{x} = \vec{0}$.

Thus the zero vector is orthogonal to everything.

Technical detail: So if the vectors are both non-zero and orthogonal, then the vectors are both perpendicular and orthogonal. But of course, since the zero vector is not non-zero, it cannot be perpendicular to anything, but it is orthogonal to everything.

Normal vectors

A normal vector is one which is perpendicular to all the points/vectors on a plane.

That is for any vector $\vec{a}$ on the plane, and a normal vector, $\vec{n}$, to that plane, we have:

$$ \vec{n} \cdot \vec{a} = \vec{0} $$

For example: given an equation of a plane, $Ax + By + Cz = D$, the normal vector is simply:

$$ \vec{n} = A\hat{i} + B\hat{j} + C\hat{k} $$

Orthonormal vectors

Given $V = \{ \vec{v_1}, \vec{v_2}, \dots, \vec{v_k} \}$ where:

This can be summed up as:

$$ \vec{v_i} \cdot \vec{v_j} = \begin{cases} 0 & i \neq j \\ 1 & i = j \end{cases} $$

This is an orthonormal set. The term comes from the fact that these vectors are all
orthogonal to each other, and they have all been normalized.

Additional vector operations

These vector operations are less common, but included for reference.

Vector outer products

For the outer product, the two vectors do not need to be of the same dimension (i.e. $x \in \mathbb R^n, y \in \mathbb R^m$), and the result is a matrix instead of a scalar:

$$ x \otimes y \in \mathbb R^{n \times m} = \begin{bmatrix} x_1y_1 & \cdots & x_1y_m \\ \vdots & \ddots & \vdots \\ x_ny_1 & \cdots & x_ny_m \end{bmatrix} $$

Note that the outer product $x \otimes y$ is equivalent to the matrix multiplication $xy^T$.

Vector cross products

Cross products are much more limited than dot products. Dot products can be calculated for any $\mathbb R^n$. Cross products are only defined in $\mathbb R^3$.

Unlike the dot product, which results in a scalar, the cross product results in a vector which is orthogonal to the original vectors (i.e. it is orthogonal to the plane defined by the two original vectors).

$$ \vec{a} = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}, \vec{b} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} $$

$$ \vec{a} \times \vec{b} = \begin{bmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{bmatrix} $$

For example:

$$ \begin{bmatrix} 1 \\ -7 \\ 1 \end{bmatrix} \times \begin{bmatrix} 5 \\ 2 \\ 4 \end{bmatrix} = \begin{bmatrix} -7\times4 - 1\times2 \\ 1\times5 - 1\times4 \\ 1\times2 - -7\times5 \end{bmatrix} = \begin{bmatrix} -30 \\ 1 \\ 37 \end{bmatrix} $$

Linear Combinations

Parametric representations of lines

Any line in an $n$-dimensional space can be represented using vectors.

Say you have a vector $\vec{v}$ and a set $S$ consisting of all scalar multiplications of that vector (where the scalar $c$ is any real number):

$$ S = \{ c\vec{v} \,|\, c \in \mathbb R \} $$

This set $S$ represents a line, since multiplying a vector with scalars does not changes its direction, only its magnitudes, so that set of vectors covers the entirety of the line.

A few of the infinite scalar multiplications which define the line.
A few of the infinite scalar multiplications which define the line.

But that line is around the origin. If you wanted to shift it, you need only to add a vector, which we'll call $\vec{x}$. So we could define a line as:

$$ L = \{ \vec{x} + c\vec{v} \,|\, c \in \mathbb R \} $$

Calculating the intersecting line for two vectors.
Calculating the intersecting line for two vectors.

For example: say you are given two vectors:

$$ \vec{a} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} ,\; \vec{b} = \begin{bmatrix} 0 \\ 3 \end{bmatrix} $$

Say you want to find the line that goes through them. First you need to find the vector along that intersecting line, which is just $\vec{b} - \vec{a}$.

Although in standard form, that vector originates at the origin.

Thus you still need to shift it by finding the appropriate vector $\vec{x}$ to add to it. But as you can probably see, we can use our $\vec{a}$ to shift it, giving us:

$$ L = \{ \vec{a} + c(\vec{b} - \vec{a}) \,|\, c \in \mathbb R \} $$

And this works for any arbitrary $n$ dimensions! (Although in other spaces, this wouldn't really be a "line". In $\mathbb R^3$ space, for instance, this would define a plane.)

You can convert this form to a parametric equation, where the equation for a dimension of the vector $a_i$ looks like:

$$ a_i + (b_i - a_i)c $$

Say you are in $\mathbb R^2$ so, you might have:

$$ L = \left\lbrace \begin{bmatrix} 0 \\ 3 \end{bmatrix} + c \begin{bmatrix} -2 \\ 2 \end{bmatrix} \,|\, c \in \mathbb R \right\rbrace $$

you can write it as the following parametric equation:

$$ \begin{aligned} x &= 0 + -2c = -2c \\ y &= 3 + 2c = 2c + 3 \end{aligned} $$

Linear combinations

Say you have the following vectors in $\mathbb R^m$ :

$$ \vec{v_1}, \vec{v_2}, \dotsc, \vec{v_n} $$

A linear combination is just some sum of the combination of these vectors, scaled by arbitrary constants ($c_1 \to c_n \in \mathbb R$):

$$ c_1\vec{v_1} + c_2\vec{v_2}, + \dotsc + c_n\vec{v_n} $$

For example:

$$ \vec{a} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} ,\; \vec{b} = \begin{bmatrix} 0 \\ 3 \end{bmatrix} $$

A linear combination would be:

$$ 0\vec{a} + 0\vec{b} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

Any vector in the space $\mathbb R^2$ can represented by some linear combination of these two vectors.


The set of all linear combinations for some vectors is called the span.

The span of some vectors can define an entire space. For instance, using our previously-defined vectors:

$$ \Span(\vec{a}, \vec{b}) = \mathbb R^2 $$

But this is not always true for the span of any arbitrary set of vectors. For instance, this does not represent all the vectors in $\mathbb R^2$ :

$$ \Span(\begin{bmatrix} -2 \\ -2 \end{bmatrix}, \begin{bmatrix} 2 \\ 2 \end{bmatrix}) $$

These two vectors are collinear (that is, they lie along the same line), so combinations of them will only yield other vectors along that line.

As another example, the span of the zero vector, $\Span(\vec{0})$, cannot represent all vectors in a space.

Formally, the span is defined as:

$$ \Span(\vec{v_1}, \vec{v_2}, \dotsc, \vec{v_n}) = \{ c_1\vec{v_1} + c_2\vec{v_2}, + \dotsc + c_n\vec{v_n} \,|\, c_i \in \mathbb R \, \forall \, 1 \le i \le n \} $$

Linear independence

The set of vectors in the previous collinear example:

$$ \left\lbrace \begin{bmatrix} -2 \\ -2 \end{bmatrix}, \begin{bmatrix} 2 \\ 2 \end{bmatrix} \right\rbrace $$

are called a linearly dependent set, which means that some vector in the set can be represented as the linear combination of some of the other vectors in the set.

In this example, we could represent $\begin{bmatrix} -2 , -2 \end{bmatrix}$ using the linear combination of the other vector, i.e. $-1\begin{bmatrix} 2 , 2 \end{bmatrix}$.

You can think of a linearly dependent set as one that contains a redundant vector - one that doesn't add any more information to the set.

As another example:

$$ \left\lbrace \begin{bmatrix} 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 7 \\ 2 \end{bmatrix}, \begin{bmatrix} 9 \\ 5 \end{bmatrix} \right\rbrace $$

is linearly dependent because $\vec{v_1} + \vec{v_2} = \vec{v_3}$.

Naturally, a set that is not linearly dependent is called a linearly independent set.

For a more formal definition of linear dependence, a set of vectors:

$$ S = \{ \vec{v_1}, \vec{v_2}, \dotsc, \vec{v_n} \} $$

is linearly dependent iff (if and only if)

$$ c_1\vec{v_1} + c_2\vec{v_2} + \dotsc + c_n\vec{v_n} = \vec{0} = \begin{bmatrix} 0 \\ \vdots \\ 0 \end{bmatrix} $$

for some $c_i$'s where at least one is non-zero.

To put the previous examples in context, if you can show that at least one of the vectors can be described by the linear combination of the other vectors in the set, that is:

$$ \vec{v_1} = a_2\vec{v_2} + a_3\vec{v_3} + \dots + a_n\vec{v_n} $$

then you have a linearly dependent set because that can be reduced to show:

$$ \vec{0} = -1\vec{v_1} + a_2\vec{v_2} + a_3\vec{v_3} + \dots + a_n\vec{v_n} $$

Thus you can calculate the zero vector as a linear combination of the vectors where at least one constant is non-zero, which satifies the definition for linear dependence.

So then a set is linearly independent if, to calculate the zero vector as a linear combination of the vectors, the coefficients must all be zero.

Going back to spans, the span of set of size $n$ which is linearly independent can describe that set's entire space (e.g. $\mathbb R^n$).

An example problem:

Say you have the set:

$$ S = \left\lbrace \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} \right\rbrace $$

and you want to know:

For the first question, you want to see if any linear combination of the set yields any arbitrary vector in $\mathbb R^3$ :

$$ c_1\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}, c_2\begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}, c_3\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} $$

You can distribute the coefficients:

$$ \begin{bmatrix} 1c_1 \\ -1c_1 \\ 2c_1 \end{bmatrix}, \begin{bmatrix} 2c_2 \\ 1c_2 \\ 3c_2 \end{bmatrix}, \begin{bmatrix} -1c_3 \\ 0c_3 \\ 2c_3 \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} $$

So you can break that out into a system of equations:

$$ \begin{aligned} c_1 + 2c_2 - c_3 &= a \\ -c_1 + c_2 + 0 &= b \\ 2c_1 + 3c_2 + 2c_3 &= c \end{aligned} $$

And solve it, which gives you:

$$ \begin{aligned} c_3 &= \frac{1}{11}(3c - 5a + b) \\ c_2 &= \frac{1}{3}(b + a + c_3) \\ c_1 &= a - 2c_2 + c_3 \end{aligned} $$

So it looks like you can get these coefficients from any $a, b, c$, so we can say $\Span(S) = \mathbb R^3$.

For the second question, we want to see if all of the coefficients have to be non-zero for this to be true:

$$ c_1\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}, c_2\begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}, c_3\begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

We can just reuse the previous equations we derived for the coefficients, substituting $a=0, b=0, c=0$, which gives us:

$$ c_1 = c_2 = c_3 = 0 $$

So we know this set is linearly independent.


A matrix can, in some sense, be thought of as a vector of vectors.

The notation $m \times n$ in terms of matrices mean there are $m$ rows and $n$ columns.

So that matrix would look like:

$$ \mathbf A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{bmatrix} $$

A matrix of these dimensions may also be notated $\mathbb R^{m \times n}$ to indicate its membership in that set.

We refer to the entry in the $i$th row and $j$th column with the notation $\mathbf A_{ij}$.

Matrix operations

Matrix addition

Matrices must have the same dimensions in order to be added (or subtracted).

$$ \mathbf A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{bmatrix}, \mathbf B = \begin{bmatrix} b_{11} & b_{12} & \dots & b_{1n} \\ b_{21} & b_{22} & \dots & b_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ b_{m1} & b_{m2} & \dots & b_{mn} \end{bmatrix} $$

$$ \mathbf A + \mathbf B = \begin{bmatrix} a_{11} + b_{11} & a_{12} + b_{12} & \dots & a_{1n} + b_{1n} \\ a_{21} + b_{21} & a_{22} + b_{22} & \dots & a_{2n} + b_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{m1} + b_{m1} & a_{m2} + b_{m2} & \dots & a_{mn} + b_{mn} \end{bmatrix} $$

$$ \mathbf A + \mathbf B = \mathbf B + \mathbf A $$

Matrix-scalar multiplication

Just distribute the scalar:

$$ \mathbf A = \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{bmatrix}, c\mathbf A = \begin{bmatrix} ca_{11} & ca_{12} & \dots & ca_{1n} \\ ca_{21} & ca_{22} & \dots & ca_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ ca_{m1} & ca_{m2} & \dots & ca_{mn} \end{bmatrix} $$

Matrix-vector products

To multiply a $m \times n$ matrix with a vector, the vector must have $n$ components (that is, the same number of components as there are columns in the matrix, i.e. $\vec{x} \in \mathbb R^n$):

$$ \vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} $$

The product would be:

$$ \mathbf A \vec{x} = \begin{bmatrix} a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n \\ a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n \end{bmatrix} $$

This results in a $m \times 1$ matrix.

Matrix-vector products as linear combinations

If you interpret each column in a matrix $\mathbf A$ as its own vector $\vec{v_i}$, such that:

$$ \mathbf A = \begin{bmatrix} \vec{v_1} & \vec{v_2} & \dots & \vec{v_n} \end{bmatrix} $$

Then the product of a matrix and vector can be rewritten simply as a linear combination of those vectors:

$$ \mathbf A \vec{x} = x_1\vec{v_1} + x_2\vec{v_2} + \dots + x_n\vec{v_n} $$

Matrix-vector products as linear transformations

A matrix-vector product can also be seen as a linear transformation. You can describe it as a transformation:

$$ \begin{aligned} \mathbf A &= \begin{bmatrix} \vec{v_1} & \vec{v_2} & \dots & \vec{v_n} \end{bmatrix} \\ T: \mathbb R^n &\to \mathbb R^m \\ T(\vec{x}) &= \mathbf A \vec{x} \end{aligned} $$

It satisfies the conditions for a linear transformation (not shown here), so a matrix-vector product is always a linear transformation.

Just to be clear: the transformation of a vector can always be expressed as that vector's product with some matrix; that matrix is referred to as the transformation matrix.

So in the equations above, $\mathbf A$ is the transformation matrix.

To reiterate:

Matrix-matrix products

To multiply two matrices, one must have the same number of columns as the other has rows. That is, you can only multiply an $m \times n$ matrix with an $n \times p$ matrix. The resulting matrix will be of $m \times p$ dimensions.

That is, if $A \in \mathbb R^{m \times n}, B \in \mathbb R^{n \times p}$, then $C = AB \in \mathbb R^{m \times p}$.

The resulting matrix is defined as such:

$$ C_{ij} = \sum^{n}_{k=1} A_{ik} B_{kj} $$

You can break the terms out into individual matrix-vector products. Then you combine the resulting vectors to get the final matrix.

More formally, the $i$th column of the resulting product matrix is obtained by multiplying $\mathbf A$ with the $i$th column of $\mathbf B$ for $i=1,2,\dots,k$.

$$ \begin{bmatrix} 1 & 3 & 2 \\ 4 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 0 & 1 \\ 5 & 2 \end{bmatrix} $$

The product would be:

$$ \begin{aligned} \begin{bmatrix} 1 & 3 & 2 \\ 4 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 \\ 0 \\ 5 \end{bmatrix} &= \begin{bmatrix} 11 \\ 9 \end{bmatrix} \\ \begin{bmatrix} 1 & 3 & 2 \\ 4 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} &= \begin{bmatrix} 10 \\ 14 \end{bmatrix} \\ \begin{bmatrix} 1 & 3 & 2 \\ 4 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 0 & 1 \\ 5 & 2 \end{bmatrix} &= \begin{bmatrix} 11 & 10 \\ 9 & 14 \end{bmatrix} \end{aligned} $$

Properties of matrix multiplication

Matrix multiplication is not commutative. That is, for matrices $\mathbf A$ and $\mathbf B$, in general $\mathbf A \times \mathbf B \neq \mathbf B \times \mathbf A$. They may not even be of the same dimension.

Matrix multiplication is associative. For example, for matrices $\mathbf A, \mathbf B, \mathbf C$, we can say that:

$$ \mathbf A \times \mathbf B \times \mathbf C = \mathbf A \times (\mathbf B \times \mathbf C) = (\mathbf A \times \mathbf B) \times \mathbf C $$

There is also an identity matrix $\mathbf I$. For any matrix $\mathbf A$, we can say that:

$$ \mathbf A \times \mathbf I = \mathbf I \times \mathbf A = \mathbf A $$

Hadamard product

The Hadamard product, sometimes called the element-wise product, is another way of multiplying matrices, but only for matrices of the same size. It is usually denoted with $\odot$. It is simply:

$$ (A \odot B)_{n,m} = A_{n,m}B_{n,m} $$

It returns a matrix of the same size as the input matrices.

Haramard multiplication has the following propertise:

The identity matrix

The identity matrix is an $n \times n$ matrix where every component is 0, except for those along the diagonal:

$$ \mathbf I_n = \begin{bmatrix} 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 1 \end{bmatrix} $$

When you multiply the identity matrix by any vector:

$$ \mathbf I_n \vec{x} \, | \, \vec{x} \in \mathbb R^n = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \vec{x} $$

That is, a vector multiplied by the identity matrix equals itself.

Diagonal matrices

A diagonal matrix is a matrix where all non-diagonal elements are 0, typically denoted $\Diag(x_1, x_2, \dots, x_n)$, where

$$ D_{ij} = \begin{cases} d_i & i=j \\ 0 & i \neq j \end{cases} $$

So the identity matrix is $I = \Diag(1, 1, \dots, 1)$.

Triangular matrices

We say that a matrix is upper-triangular if all of its elements below the diagonal are zero.

Similarly, a matrix is lower-triangular if all of its elements above the diagonal are zero.

A matrix is diagonal if it is both upper-triangular and lower-triangular.

Some properties of matrices

Associative property

$$ (AB)C = A(BC) $$

i.e. it doesn't matter where the parentheses are.

This applies to compositions as well:

$$ (h \circ f) \circ g = h \circ (f \circ g) $$

Distributive property

$$ \begin{aligned} A(B+C) &= AB + AC \\ (B+C)A &= BA + CA \end{aligned} $$

Matrix inverses

If $\mathbf A$ is an $m \times m$ matrix, and if it has an inverse, then:

$$ \mathbf A \mathbf A^{-1} = \mathbf A^{-1} \mathbf A = \mathbf I $$

Only square matrices can have inverses. An inverse does not exist for all square matrices, but those that have one are called invertible or non-singular, otherwise they are non-invertible or singular.

The inverse exists if and only if $A$ is full rank.

The invertible matrices $A,B \in \mathbb R^{n \times n}$ have the following properties:


$A^{\dagger}$ is a pseudo-inverse (sometimes called a Moore-Penrose inverse) of $A$, which may be non-square, if the following are satisfied:

A pseudo-inverse exists and is unique for any matrix $A$. If $A$ is invertible, $A^{-1} = A^{\dagger}$.

Matrix determinants

The determinant of a square matrix $A \in \mathbb R^{n \times n}$ is a function $\Det : \mathbb R^{n \times n} \to \mathbb R$, denoted $|A|$, $\Det(A)$, or sometimes with the parentheses dropped, $\Det A$.

A more intuitive interpretation:

Say we are in a 2D space and we have some shape. It has some area. Then we apply transformations to that space. The determinant describes how that shape's area has been scaled as a result of the transformation.

This can be extended to 3D, replacing "area" with "volume".

With this interpretation it's clear that a determinant of 0 means scaling area or volume to 0, which indicates that space has been "compressed" to a line or a point (or a plane in the case of 3D).

Inverse and determinant for a $2 \times 2$ matrix

Say you have the matrix:

$$ \mathbf A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$

You can calculate the inverse of this matrix as:

$$ \mathbf A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} $$

Note that $\mathbf A^{-1}$ is undefined if $ad-bc=0$, which means that $\mathbf A$ is not invertible.


The denominator $ad-bc$ is called the determinant. It is notated as:

$$ \Det(\mathbf A) = |\mathbf A| = ad-db $$

Inverse and determinant for an $n \times n$ matrix

Say we have an $n \times n$ matrix $\mathbf A$.

A submatrix of $\mathbf A_{ij}$ is an $(n-1)\times(n-1)$ matrix constructed from $\mathbf A$ by ignoring the $i^{th}$ row and the $j^{th}$ column of $\mathbf A$, which we denote by $\mathbf A_{\lnot i, \lnot j}$.

You can calculate the determinant of an $n \times n$ matrix $\mathbf A$ by using some $i^{th}$ row of $\mathbf A$, where $1 \le i \le n$:

$$ \Det(\mathbf A) = \sum_{j=1}^n (-1)^{i+j}a_{ij} \Det(\mathbf A_{\lnot i,\lnot j}) $$

All the $\Det(\mathbf A_{ij})$ eventually reduce to the determinant of a $2 \times 2$ matrix.

Scalar multiplication of determinants

For an $n \times n$ matrix $\mathbf A$,

$$ \Det(k\mathbf A) = k^n \Det(\mathbf A) $$

Determinant of diagonal or triangular matrix

The determinant of a diagonal or triangular matrix is simply the product of the elements along its diagonal.

Properties of determinants

- For $A \in \mathbb R^{n \times n}, t \in \mathbb R$, multiplying a single row by the scalar $t$ yields a new matrix $B$, for which $ B = t A $.
- For $A,B \in \mathbb R^{n \times n}$, $ AB = A B
- For $A,B \in \mathbb R^{n \times n}$, $ A = 0$ if $A$ is singular (i.e. non-invertible).
- For $A,B \in \mathbb R^{n \times n}$, $ A ^{-1} = \frac{1}{ A }$ if $A$ is non-singular (i.e. invertible).
- For $A,B \in \mathbb R^{n \times n}$, if two rows of $A$ are swapped to produce $B$, then $\Det(A) = -\Det(B)$
- The determinant of a matrix $A \in \mathbb R^{n \times n}$ is non-zero if and only if it has full rank; this also means you can check if a $A$ is invertible by checking that its determinant is non-zero.

Transpose of a matrix

The transpose of a matrix $\mathbf A$ is that matrix with its columns and rows swapped, denoted $\mathbf A^T$.

More formally, let $\mathbf A$ be an $m \times n$ matrix, and let $\mathbf B = \mathbf A^T$. Then $\mathbf B$ is an $n \times m$ matrix, and $B_{ij} = A_{ji}$.

Symmetric matrices

A square matrix $A \in \mathbb R^{n \times n}$ is symmetric if $A = A^T$.

It is anti-symmetric if $A = -A^T$.

For any square matrix $A \in \mathbb R^{n \times n}$, the matrix $A + A^T$ is symmetric and the matrix $A - A^T$ is anti-symmetric. Thus any such $A$ can be represented as a sum of a symmetric and an anti-symmetric matrix:

$$ A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T) $$

Symmetric matrices have many nice properties.

The set of all symmetric matrices of dimension $n$ is often denoted as $\mathbb S^n$, so you can denote a symmetric $n \times n$ matrix $A$ as $A \in \mathbb S^n$.

The quadratic form

Given a square matrix $A \in \mathbb R^{n \times n}$ and a vector $x \in \mathbb R$, the scalar value $x^TAx$ is called a quadratic form:

$$ x^TAx = \sum^n_{i=1} \sum^n_{j=1} A_{ij}x_ix_j $$

Here $A$ is typically assumed to be symmetric.

Types of symmetric matrices

Given a symmetric matrix $A \in \mathbb S^n$...

Some other properties of note:

Essentially, "positive semidefinite" is to matrices as "non-negative" is to scalar values (and "positive definite" as "positive" is to scalar values).

The Trace

The trace of a square matrix $A \in \mathbb R^{n \times n}$ is denoted $\Tr(A)$ and is the sum of the diagonal elements in the matrix:

$$ \Tr(A) = \sum^n_{i=1} A_{ii} $$

The trace has the following properties:

Orthogonal matrix

Say we have a $n \times k$ matrix $\mathbf C$, whose column rows form an orthonormal set.

If $k=n$ then $\mathbf C$ is a square matrix ($n \times n$) and since $\mathbf C$'s columns are linearly independent, $\mathbf C$ is invertible.

For an orthonormal matrix:

$$ \begin{aligned}[c] \mathbf C^T \mathbf C &= \mathbf I_n \\ \mathbf C^{-1} \mathbf C &= \mathbf I_n \end{aligned} \qquad \begin{aligned}[c] \therefore \mathbf C^T = \mathbf C^{-1} \end{aligned} $$

When $\mathbf C$ is an $n \times n$ matrix (i.e. square) whose columns form an orthonormal set, we say that $\mathbf C$ is an orthogonal matrix.

Orthogonal matrices have the property of $C^TC = I = CC^T$.

Orthogonal matrices also have the property that operating on a vector with an orthogonal matrix will not change its Euclidean norm, i.e. $||Cx||_2 = ||x||_2$ for any $x \in \mathbb R^n$.

Orthogonal matrices preserve angles and lengths

For an orthogonal matrix $\mathbf C$, when you multiply $\mathbf C$ by some vector, the length and angle of the vector is preserved:

$$ \begin{aligned} ||\vec{x}|| &= ||\mathbf C \vec{x}|| \\ \cos\theta &= \cos\theta_{\mathbf C} \end{aligned} $$


The classical adjoint, often just called the adjoint of a matrix $A \in \mathbb R^{n \times n}$ is denoted $\Adj(A)$ and defined as:

$$ \Adj(A)_{ij} = (-1)^{i+j}|A_{\lnot j, \lnot i}| $$

Note that the indices are switched in $A_{\lnot j, \lnot i}$.


Say we have set of vectors $V$ which is a subset of $\mathbb R^n$, that is, every vector in the set has $n$ components.

A subspace
A subspace

$V$ is a linear subspace of $\mathbb R^n$ if


Say we have the set of vectors:

$$ S = \left\lbrace \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \in \mathbb R^2 \,:\, x_1 \ge 0 \right\rbrace $$

which is the shaded area below.

Is the shaded set of vectors $S$ a subspace of $\mathbb R^2$?
Is the shaded set of vectors $S$ a subspace of $\mathbb R^2$?

Is $S$ a subspace of $\mathbb R^2$?

So no, this set is not a subspace of $\mathbb R^2$.

Spans and subspaces

Let's say we have the set:

$$ U = \Span(\vec{v_1}, \vec{v_2}, \vec{v_3}) $$

where each vector has $n$ components.
Is this a valid subspace of $\mathbb R^n$?

Since the span represents all the linear combinations of those vectors, we can define an arbitrary vector in the set as:

$$ \vec{x} = c_1\vec{v_1} + c_2\vec{v_2} + c_3\vec{v_3} $$

Basis of a subspace

If we have a subspace $V = \Span(S)$ where the set of vectors $S = \vec{v_1}, \vec{v_2}, \dots, \vec{v_n}$ is linearly independent, then we can say that $S$ is a basis for $V$.

A set $S$ is the basis for a subspace $V$ if $S$ is linearly independent and its span defines $V$. In other words, the basis is the minimum set of vectors that spans the subspace that it is a basis of.

All bases for a subspace will have the same number of elements.

Intuitively, the basis of a subspace is a set of vectors that can be linearly combined to describe any vector in that subspace. For example, the vectors [0,1] and [1,0] form a basis for $\mathbb R^2$.

Dimension of a subspace

The dimension of a subspace is the number of elements in a basis for that subspace.

Nullspace of a matrix

Say we have:

$$ \mathbf A \vec{x} = \vec{0} $$

If you have a set $N$ of all $x \in \mathbb R^n$ that satisfies this equation, do you have a valid subspace?

Of course if $\vec{x} = \vec{0}$ this equation is satisfied. So we know the zero vector is part of this set (which is a requirement for a valid subspace).

The other two properties (closure under multiplication and addition) necessary for a subspace also hold:

$$ \begin{aligned} A(\vec{v_1} + \vec{v_2}) &= A\vec{v_1} + A\vec{v_2} = \vec{0} \\ A(c\vec{v_1}) &= \vec{0} \end{aligned} $$

and of course $\vec{0}$ is in the set $N$.

So yes, the set $N$ is a valid subspace, and it is a special subspace: the nullspace of $\mathbf A$, notated:

$$ N(\mathbf A) $$

That is, the nullspace for a matrix $\mathbf A$ is the subspace described by the set of vectors which yields the zero vector which multiplied by $\mathbf A$, that is, the set of vectors which are the solutions for $\vec{x}$ in:

$$ \mathbf A \vec{x} = \vec{0} $$

Or, more formally, if $\mathbf A$ is an $m \times n$ matrix:

$$ N(\mathbf A) = \{ \vec{x} \in \mathbb R^n \, | \, \mathbf A \vec{x} = \vec{0} \} $$

The nullspace for a matrix $A$ may be notated $\mathcal N(A)$.

To put nullspace (or "kernel") another way, it is space of all vectors that map to the zero vector after applying the transformation the matrix represents.

Nullspace and linear independence

If you take each column in a matrix $\mathbf A$ as a vector $\vec{v_i}$, that set of vectors is linearly independent if the nullspace of $\mathbf A$ consists of only the zero vector. That is, if:

$$ N(\mathbf A) = \{\vec{0}\} $$

The intuition behind this is because, if the linear combination of a set of vectors can only equal the zero vector if all of its coefficients are zero (that is, its coefficients are components of the zero vector), then it is linearly independent:

$$ x_1\vec{v_1} + x_2\vec{v_2} + \dots + x_n\vec{v_n} = \vec{0} \; \text{iff} \; \vec{x} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} $$


The nullity of a nullspace is its dimension, that is, it is the number of elements in a basis for that nullspace.

$$ \Dim(N(\mathbf A)) = \text{nullity}(N(\mathbf A)) $$

Left nullspace

The left nullspace of a matrix $\mathbf A$ is the nullspace of its transpose, that is $N(\mathbf A^T)$ :

$$ N(\mathbf A^T) = \{ \vec{x} \, | \, \vec{x}^T \mathbf A = \vec{0}^T \} $$


Again, a matrix can be represented as a set of column vectors.
The columnspace of a matrix (also called the range of the matrix) is all the linear combinations (i.e. the span) of these column vectors:

$$ \begin{aligned} \mathbf A &= \begin{bmatrix} \vec{v_1} & \vec{v_2} & \dots & \vec{v_n} \end{bmatrix} \\ C(\mathbf A) &= \Span(\vec{v_1}, \vec{v_2}, \dots, \vec{v_n}) \end{aligned} $$

Because any span is a valid subspace, the columnspace of a matrix is a valid subspace.

So if you expand out the matrix-vector product, you'll see that every matrix-vector product is within that matrix's columnspace:

$$ \begin{aligned} \{ \mathbf A \vec{x} \, &| \, \vec{x} \in \mathbb R^n \} \\ \mathbf A \vec{x} &= x_1\vec{v_1} + x_2\vec{v_2} + \dots + x_n\vec{v_n} \\ \mathbf A \vec{x} &= C(\mathbf A) \end{aligned} $$

That is, for any vector in the space $\mathbb R^n$, multiplying the matrix by it just yields another linear combination of that matrix's column vectors. Therefore it is also in the columnspace.

The columnspace (range) for a matrix $A$ may be notated $\mathcal R(A)$.

Rank of a columnspace

The column rank of a columnspace is its dimension, that is, it is the number of elements in a basis for that columnspace (i.e. the largest number of columns of the matrix which constitute a linearly independent set):

$$ \Dim(C(\mathbf A)) = \Rank(C(\mathbf A)) $$


The rowspace of a matrix $\mathbf A$ is the columnspace of $\mathbf A^T$, i.e. $C(\mathbf A^T)$.

The row rank of a matrix is similarly the number of elements in a basis for that rowspace.


Note that for any matrix $A$, the column rank and the row rank are equal, so they are typically just referred to as $\Rank(A)$.

The rank has some properties:

The rank of a transformation refers to the number of dimensions in the output.

A matrix is full-rank if it has rank equal to the number of dimensions in its originating space. I.e. they represent a transformation that preserves the dimensionality (it does not collapse it to a lower dimension space).

The standard basis

The set of column vectors in an identity matrix $\mathbf I_n$ is known as the standard basis for $\mathbb R^n$.

Each of those column vectors is notated $\vec{e_i}$. E.g., in an identity matrix, the column vector:

$$ \begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = \vec{e_1} $$

For a transformation $T(\vec{x})$, its transformation matrix $\mathbf A$ can be expressed as:

$$ \mathbf A = \begin{bmatrix} T(\vec{e_1}) & T(\vec{e_2}) & \dots & T(\vec{e_n}) \end{bmatrix} $$

Orthogonal compliments

Given that $V$ is some subspace of $\mathbb R^n$, the orthogonal compliment of $V$, notated $V^{\perp}$ :

$$ V^{\perp} = \{ \vec{x} \in \mathbb R^n \, | \, \vec{x} \cdot \vec{v} = 0 \, \forall \, \vec{v} \in V \} $$

That is, the orthogonal compliment of a subspace $V$ is the set of all vectors where the dot product of each vector with each vector from $V$ is 0, that is where all vectors in the set are orthogonal to all vectors in $V$.

$V^{\perp}$ is a subspace (proof omitted).

Columnspaces, nullspaces, and transposes

$C(\mathbf A)$ is the orthogonal compliment to $N(\mathbf A^T)$, and vice versa:

$$ \begin{aligned} N(\mathbf A^T) &= C(\mathbf A)^{\perp} \\ N(\mathbf A^T)^{\perp} &= C(\mathbf A) \end{aligned} $$

$C(\mathbf A^T)$ is the orthogonal compliment to $N(\mathbf A)$, and vice versa.

$$ \begin{aligned} N(\mathbf A) &= C(\mathbf A^T)^{\perp} \\ N(\mathbf A)^{\perp} &= C(\mathbf A^T) \end{aligned} $$

As a reminder, columnspaces and nullspaces are spans, i.e. sets of linear combinations, i.e. lines, so these lines are orthogonal to each other.

Dimensionality and orthogonal compliments

For $V$, a subspace of $\mathbb R^n$ :

$$ \Dim(V) + \Dim(V^{\perp}) = n$$

(proof omitted here)

The intersection of orthogonal compliments

Since the vectors between a subspace and its orthogonal compliment are all orthogonal:

$$ V \cap V^{\perp} = \{ \vec{0} \} $$

That is, the only vector which exists both in a subspace and its orthogonal compliment is the zero vector.

Coordinates with respect to a basis

With a subspace $V$ of $\mathbb R^n$, we have $V$'s basis, $B$, as

$$ B = \{ \vec{v_1}, \vec{v_2}, \dots, \vec{v_k} \} $$

We can describe any vector $\vec{a} \in V$ as a linear combination of the vectors in its basis
$B$ :

$$ \vec{a} = c_1\vec{v_1} + c_2\vec{v_2} + \dots + c_k\vec{v_k} $$

We can take these coefficients $c_1, c_2, \dots, c_k$ as the coordinates of $\vec{a}$ with respect to $B$, notated as:

$$ [\vec{a}]_B = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_k \end{bmatrix} $$

Basically what has happened here is a new coordinate system based off of the basis $B$ is being used.


Say we have $\vec{v_1} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \vec{v_2} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$, where $B = \{\vec{v_1}, \vec{v_2}\}$ is the basis for $\mathbb R^2$.

The point $(8,7)$ in $\mathbb R^2$ is equal to $3\vec{v_1} + 2\vec{v_2}$. If we set:

$$ \vec{a} = 3\vec{v_1} + 2\vec{v_2} $$

Then we can describe $\vec{a}$ with respect to $B$ :

$$ [\vec{a}]_B = \begin{bmatrix} 3 \\ 2 \end{bmatrix} $$

which looks like:

Coordinates wrt a basis
Coordinates wrt a basis

Change of basis matrix

Given the basis:

$$ B = \{ \vec{v_1}, \vec{v_2}, \dots, \vec{v_k} \} $$


$$ [\vec{a}]_B = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_k \end{bmatrix} $$

say there is some $n \times k$ matrix where the column vectors are the basis vectors:

$$ \mathbf C = [\vec{v_1}, \vec{v_2}, \dots, \vec{v_k}] $$

We can do:

$$ \mathbf C [\vec{a}]_B = \vec{a} $$

The matrix $\mathbf C$ is known as the change of basis matrix and allows us to get $\vec{a}$ in standard coordinates.

Invertible change of basis matrix

Given the basis of some subspace:

$$ B = \{ \vec{v_1}, \vec{v_2}, \dots, \vec{v_k} \} $$

where $\vec{v_1}, \vec{v_2}, \dots, \vec{v_k} \in \mathbb R^n$, and we have a change of basis matrix:

$$ \mathbf C = [\vec{v_1}, \vec{v_2}, \dots, \vec{v_k}] $$


Under these assumptions:


$$ [\vec{a}]_B = C^{-1}\vec{a} $$

Transformation matrix with respect to a basis

Say we have a linear transformation $T: \mathbb R^n \to \mathbb R^n$, which we can express as $T(\vec{x}) = \mathbf A \vec{x}$. This is with respect to the standard basis; we can say $\mathbf A$ is the transformation for $T$ with respect to the standard basis.

Say we have another basis $B = \{ \vec{v_1}, \vec{v_2}, \dots, \vec{v_n} \}$ for $\mathbb R^n$, that is, it is a basis for for $\mathbb R^n$.

We could write:

$$ [T(\vec{x})]_B = \mathbf D [\vec{x}]_B $$

and we call $\mathbf D$ the transformation matrix for $T$ with respect to the basis $B$.

Then we have (proof omitted):

$$ \mathbf D = \mathbf C^{-1} \mathbf A \mathbf C $$


Orthonormal bases

If $B$ is an orthonormal set, it is linearly independent, and thus it could be a basis. If $B$ is a basis, then it is an orthonormal basis.

Coordinates with respect to orthonormal bases

Orthonormal bases make good coordinate systems - it is much easier to find $[\vec{x}]_B$ if $B$ is an orthonormal basis. It is just:

$$ [\vec{x}]_B = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_k \end{bmatrix} = \begin{bmatrix} \vec{v_1} \cdot \vec{x} \\ \vec{v_2} \cdot \vec{x} \\ \vdots \\ \vec{v_k} \cdot \vec{x} \end{bmatrix} $$

Note that the standard basis for $\mathbb R^n$ is an orthonormal basis.


A transformation is just a function which operates on vectors, which, instead of using $f$, is usually denoted $T$.

Linear transformations

A linear transformation is a transformation:

$$ T: \mathbb R^n \to \mathbb R^m $$

where we can take two vectors $\vec{a}, \vec{b} \in \mathbb R^n$ and the following conditions are satisfied:

$$ \begin{aligned} T(\vec{a} + \vec{b}) &= T(\vec{a}) + T(\vec{b}) \\ T(c\vec{a}) &= cT(\vec{a}) \end{aligned} $$

Put another way, a linear transformation is a transformation in which lines are preserved (they don't become curves) and the origin remains at the origin. This can be thought of as transforming space such that grid lines remain parallel and evenly-spaced.

A linear transformation of a space can be described in terms of transformations of the space's basis vectors, e.g. $\hat i, \hat j$. For example, if the basis vectors $\hat i, \hat j$ end up at $[a,c], [b,d]$ respectively, an arbitrary vector $[x,y]$ would be transformed to:

$$ x \begin{bmatrix}a \\ c \end{bmatrix} + y \begin{bmatrix}b \\ d \end{bmatrix} = \begin{bmatrix} ax + by \\ cx + dy \end{bmatrix} $$

which is equivalent to:

$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$

In this way, we can think of matrices as representing a transformation of space and the transformation itself as a product with that matrix.

Extending this further, you can think of matrix multiplication as a composition of transformations - each matrix represents one transformation; the resulting matrix product is a composition of those transformations.

The matrix does not have to be square; i.e. it does not have to share dimensionality (in terms of the matrix's rows) with the space it's being applied to. The resulting transformation will have different dimensions.

For example, a 3x2 matrix will transform a 2D space to a 3D space.

Linear transformation examples

These examples are all in $\mathbb R^2$ since it's easier to visualize. But you can scale them up to any $\mathbb R^n$.

An example of reflection.
An example of reflection.

To get from the triangle on the left and reflect it over the $y$-axis to get the triangle on the right, all you're doing is changing the sign of all the $x$ values.

So a transformation would look like:

$$ T(\begin{bmatrix} x \\ y \end{bmatrix}) = \begin{bmatrix} -x \\ y \end{bmatrix} $$


Say you want to double the size of the triangle instead of flipping it. You'd just scale up all of its values:

$$ T(\begin{bmatrix} x \\ y \end{bmatrix}) = \begin{bmatrix} 2x \\ 2y \end{bmatrix} $$

Compositions of linear transformation

The composition of linear transformations $S(\vec{x}) = A\vec{x}$ and $T(\vec{x}) = B\vec{x}$ is denoted:

$$ T \circ S(\vec{x}) = T(S(\vec{x})) $$

This is read: "the composition of $T$ with $S$".

If $T: Y \to Z$ and $S: X \to Y$, then $T \circ S: X \to Z$.

A composition of linear transformations is also a linear transformation (proof omitted here). Because of this, this composition can also be expressed:

$$ T \circ S(\vec{X}) = C\vec{x} $$

Where $C = BA$ (proof omitted), so:

$$ T \circ S(\vec{X}) = BA\vec{x} $$


The kernel of T, denoted $\Ker(T)$, is all of the vectors in the domain such that the transformation of those vectors is equal to the zero vector:

$$ \Ker(T) = \{ \vec{x} \in \mathbb R^n \, | \, T(\vec{x}) = \{ \vec{0} \} \} $$

You may notice that, because $T(\vec{x}) = \mathbf A \vec{x}$,

$$ \Ker(T) = N(\mathbf A) $$

That is, the kernel of the transformation is the same as the nullspace of the transformation matrix.


Image of a subset of a domain

When you pass a set of vectors (i.e. a subset of a domain $\mathbb R^n$) through a transformation, the result is called the image of the set under the transformation. E.g. $T(S)$ is the image of $S$ under $T$.

For example, say we have some vectors which define the triangle on the left. When a transformation is applied to that set, the image is the result on the right.

Example: Image of a triangle.
Example: Image of a triangle.

Another example: if we have a transformation $T: X \to Y$ and $A$ which is a subset of $T$, then $T(A)$ is the image of $A$ under $T$, which is equivalent to the set of transformations for each vector in $A$ :

$$ T(A) = \{ T(\vec{x}) \in Y \, | \, \vec{x} \in A \} $$

$T: X \to Y$, where $A \subseteq X$.
$T: X \to Y$, where $A \subseteq X$.

Images describe surjective functions, that is, a surjective function $f : X \to Y$ can also be written:

$$ \Img(f) = Y $$

since the image of the transformation encompasses the entire codomain $Y$.

Image of a subspace

The image of a subspace under a transformation is also a subspace. That is, if $V$ is a subspace, $T(V)$ is also a subspace.

Image of a transformation

If, instead of a subset or subspace, you take the transformation of an entire space, i.e. $T(\mathbb R^n)$, the terminology is different: that is called the image of $T$, notated $\Img(T)$.

Because we know matrix-vector products are linear transformations:

$$ T(\vec{x}) = \mathbf A \vec{x} $$

The image of a linear transformation matrix $\mathbf A$ is equivalent to its column space, that is:

$$ \Img(T) = C(\mathbf A) $$

Preimage of a set

The preimage is the inverse image. For instance, consider a transformation mapping from the domain $X$ to the codomain $Y$ :

$$ T: X \to Y $$

And say you have a set $S$ which is a subset of $Y$. You want to find the set of values in $X$ which map to $S$, that is, the subset of $X$ for which $S$ is the image.

For a set $S$, this is notated:

$$ T^{-1}(S) $$

Note that not every point in $S$ needs to map back to $X$. That is, $S$ may contain some points for which there are no corresponding points in $X$. Because of this, the image of the preimage of $S$ is not necessarily equivalent to $S$, but we can be sure that it is at least a subset:

$$ T(T^{-1}(S)) \subseteq S $$


A projection can kind of be thought of as a "shadow" of a vector:

Here, we have the projection of $\vec{x}$ - the red vector - onto the green line $L$. The projection is the dark red vector. This example is in $\mathbb R^2$ but this works in any $\mathbb R^n$.
Here, we have the projection of $\vec{x}$ - the red vector - onto the green line $L$. The projection is the dark red vector. This example is in $\mathbb R^2$ but this works in any $\mathbb R^n$.

Alternatively, it can be thought of answering "how far does one vector go in the direction of another vector?". In the accompanying figure, the projection of $b$ onto $a$ tells us how far $b$ goes in the direction of $a$.

The projection of $\vec{x}$ onto line $L$ is notated:

$$ \Proj_L(\vec{x}) $$

More formally, a projection of a vector $\vec{x}$ onto a line $L$ is some vector in $L$ where $\vec{x} - \Proj_L(\vec{x})$ is orthogonal to $L$.

A line can be expressed as the set of all scalar multiples of a vector, i.e:

$$ L = \{ c\vec{v} \, | \, c \in \mathbb R \} $$

So we know that "some vector in $L$" can be represented as $c\vec{v}$ :

$$ \Proj_L(\vec{x}) = c\vec{v} $$

By our definition of a projection, we also know that $\vec{x} - \Proj_L(\vec{x})$ is orthogonal to $L$, which can now be rewritten as:

$$ (\vec{x} - c\vec{v}) \cdot \vec{v} = \vec{0} $$

(This is the definition of orthogonal vectors.)

Written in terms of $c$, this simplifies down to:

$$ c = \frac{\vec{x} \cdot \vec{v}}{\vec{v} \cdot \vec{v}} $$

So then we can rewrite:

$$ \Proj_L(\vec{x}) = \frac{\vec{x} \cdot \vec{v}}{\vec{v} \cdot \vec{v}}\vec{v} $$

or, better:

$$ \Proj_L(\vec{x}) = \frac{\vec{x} \cdot \vec{v}}{||\vec{v}||^2}\vec{v} $$

And you can pick whatever vector for $\vec{v}$ so long as it is part of line $L$.

However, if $\vec{v}$ is a unit vector, then the projection is simplified even further:

$$ \Proj_L(\vec{x}) = (\vec{x} \cdot \hat{u}) \hat{u} $$

Projections are linear transformations (they satisfy the requirements, proof omitted), so you can represent them as matrix-vector products:

$$ \Proj_L(\vec{x}) = \mathbf A \vec{x} $$

where the transformation matrix $\mathbf A$ is:

$$ \mathbf A = \begin{bmatrix} u_1^2 & u_2u_1 \\ u_1u_2 & u_2^2 \end{bmatrix} $$

where $u_i$ are components of the unit vector.

Also note that the length of a projection (i.e. the scalar component of the projection) is given by the dot product of the two vectors. For example, in the accompanying figure, the length of $\Proj_{\vec a}(\vec b)$ is $a \cdot b$.

Projections onto subspaces

Given that $V$ is a subspace of $\mathbb R^n$, we know that $V^{\perp}$ is also a subspace of $\mathbb R^n$, and we have a vector $\vec{x}$ such that $\vec{x} \in \mathbb R^n$, we know that $\vec{x} = \vec{v} + \vec{w}$ where $\vec{v} \in V$ and $\vec{w} \in V^{\perp}$, then:

$$ \begin{aligned} \Proj_V\vec{x} &= \vec{v} \\ \Proj_{V^{\perp}}\vec{x} &= \vec{w} \end{aligned} $$

Reminder: a projection onto a subspace is the same as a projection onto a line (a line is a subspace):

$$ \Proj_V\vec{x} = \frac{\vec{x} \cdot \vec{v}}{\vec{v}\cdot\vec{v}}\vec{v} $$


$$ \begin{aligned} V &= \Span(\vec{v}) \\ V &= \{ c\vec{v} \, | \, c \in \mathbb R \} \end{aligned} $$

So $\Proj_V \vec{x}$ is the unique vector $\vec{v} \in V$ such that $\vec{x} = \vec{v} + \vec{w}$ where $\vec{w}$ is a unique member of $V^{\perp}$.

Projection onto a subspace as a linear transform

$$ \Proj_V(\vec{x}) = \mathbf A (\mathbf A^T \mathbf A)^{-1}\mathbf A^T \vec{x} $$

where $V$ is a subspace of $\mathbb R^n$.

Note that $\mathbf A (\mathbf A^T \mathbf A)^{-1}\mathbf A^T$ is just some matrix, which we can call $\mathbf B$, so this is in the form of a linear transform, $\mathbf B \vec{x}$.

Also $\vec{v} = \Proj_V\vec{x}$, so:

$$ \vec{x} = \Proj_V\vec{x} + \vec{w} $$

where $\vec{w}$ is a unique member of $V^{\perp}$.

Projections onto subspaces with orthonormal bases

Given that $V$ is a subspace of $\mathbb R^n$ and $B = \{ \vec{v_1}, \vec{v_2}, \dots, \vec{v_k} \}$ is an orthonormal basis for $V$. We have a vector $\vec{x} \in \mathbb R^n$, so $\vec{x} = \vec{v} + \vec{w}$ where $\vec{v} \in V$ and $\vec{w} \in V^{\perp}$.

We know (see previously) that by definition:

$$ \Proj_V(\vec{x}) = \mathbf A (\mathbf A^T \mathbf A)^{-1}\mathbf A^T \vec{x} $$

which is quite complicated. It is much simpler for orthonormal bases:

$$ \Proj_V(\vec{x}) = \mathbf A \mathbf A^T \vec{x} $$

Identifying transformation properties

Determining if a transformation is surjective

A transformation $T(\vec{x}) = \mathbf A \vec{x}$ is surjective ("onto") if the column space of $\mathbf A$ equals the codomain:

$$ \Span(a_1, a_2, \dots, a_n) = C(\mathbf A) = \mathbb R^m $$

which can also be stated as:

$$ \Rank(\mathbf A) = m $$

Determining if a transformation is injective

A transformation $T(\vec{x}) = \mathbf A \vec{x}$ is injective ("one-to-one") if the the nullspace of $\mathbf A$ contains only the zero vector:

$$ N(\mathbf A) = \{ \vec{0} \} $$

which is true if the set of $\mathbf A$'s column vectors is linearly independent.

This can also be stated as:

$$ \Rank(\mathbf A) = n $$

Determining if a transformation is invertible

A transformation is invertible if it is both injective and surjective.

For a transformation to be surjective:

$$ \Rank(\mathbf A) = m $$

And for a transformation to be surjective:

$$ \Rank(\mathbf A) = n $$

Therefore for a transformation to be invertible:

$$ \Rank(\mathbf A) = m = n $$

So the transformation matrix $\mathbf A$ must be a square matrix.

Inverse transformations of linear transformations

Inverse transformations are linear transformations if the original transformation is both linear and invertible. That is, if $T$ is invertible and linear, $T^{-1}$ is linear:

$$ \begin{aligned} T^{-1}(\vec{x}) &= \mathbf A^{-1} \vec{x} \\ (T^{-1} \circ T)(\vec{x}) &= \mathbf A^{-1} \mathbf A \vec{x} = I_n \vec{x} = \mathbf A \mathbf A^{-1} \vec{x} = (T \circ T^{-1})(\vec{x}) \end{aligned} $$

Eigenvalues and Eigenvectors

Say we have a linear transformation $T: \mathbb R^n \to \mathbb R^n$ :

$$ T(\vec{v}) = \mathbf A \vec{v} = \lambda\vec{v} $$

That is, $\vec{v}$ is scaled by a transformation matrix $\lambda$.

We say that:

Eigenvectors are vectors for which matrix multiplication is equivalent to only a scalar multiplication, nothing more. $\lambda$, the eigenvalue, is the scalar that the transformation matrix $\mathbf A$ is equivalent to.

Another way to put this: given a square matrix $A \in \mathbb R^{n \times n}$, we say $\lambda \in \mathbb C$ is an eigenvalue of $A$ and $x \in \mathbb C^n$ is the corresponding eigenvector if:

$$ Ax = \lambda x, x \neq 0 $$

Note that $\mathbb C$ refers to the set of complex numbers.

So this means that multiplying $A$ by $x$ just results in a new vector which points in the same direction has $x$ but scaled by a factor $\lambda$.

For any eigenvector $x \in \mathbb C^n$ and a scalar $t \in \mathbb C$, $A(cx) = cAx = c \lambda x = \lambda(cx)$, that is, $cx$ is also an eigenvector - but when talking about "the" eigenvector associated with $\lambda$, it is assumed that the eigenvector is normalized to length 1 (though you still have the ambiguity that both $x$ and $-x$ are eigenvectors in this sense).

Eigenvalues and eigenvectors come up when maximizing some function of a matrix.

So what are our eigenvectors? What $\vec{v}$ satisfies:

$$ \mathbf A \vec{v} = \lambda\vec{v}, \vec{v} \neq 0$$

We can do:

$$ \begin{aligned} \mathbf A \vec{v} &= \lambda \vec{v} \\ \vec{0} &= \lambda \vec{v} - \mathbf A \vec{v} \end{aligned} $$

We know that $\vec{v} = \mathbf I_n \vec{v}$, so we can do:

$$ \vec{0} = \lambda \mathbf I_n \vec{v} - \mathbf A \vec{v} = (\lambda \mathbf I_n - \mathbf A)\vec{v} $$

The first term, $\lambda \mathbf I_n - \mathbf A$, is just some matrix which we can call $\mathbf B$, so we have:

$$ \vec{0} = \mathbf B \vec{v} $$

which, by our definition of nullspace, indicates that $\vec{v}$ is in the nullspace of $\mathbf B$. That is:

$$ \vec{v} \in N(\lambda \mathbf I_n - \mathbf A) $$

Properties of eigenvalues and eigenvectors

Diagonalizable matrices

All eigenvector equations can be written simultaneously as:

$$ AX = X\Lambda $$

where the columns of $X \in \mathbb R^{n \times n}$ are the eigenvectors of $A$ and $\Lambda$ is a diagonal matrix whose entries are the eigenvalues of $A$, i.e. $\Lambda = \Diag(\lambda_1, \dots, \lambda_n)$.

If the eigenvectors of $A$ are linearly independent, then the matrix $X$ will be invertible, so that $A = X \Lambda X^{-1}$. A matrix that can be written in this form is called diagonalizable.

Eigenvalues & eigenvectors of symmetric matrices

For a symmetric matrix $A \in \mathbb S^n$, the eigenvalues of $A$ are all real and the eigenvectors of $A$ are all orthonormal.

If for all of $A$'s eigenvalues $\lambda_i$...


Say we have a linear transformation $T(\vec{x}) = \mathbf A \vec{x}$. Here are some example values of $\vec{x}$ being input and the output vectors they yield (it's not important here what $\mathbf A$ actually looks like, its just to help distinguish what is and isn't an eigenvector.)


The eigenvectors that correspond to an eigenvalue $\lambda$ form the eigenspace for that $\lambda$, notated $E_{\lambda}$ :

$$ E_{\lambda} = N(\lambda \mathbf I_n - \mathbf A) $$


Say we have an $n \times n$ matrix $\mathbf A$. An eigenbasis is a basis for $\mathbb R^n$ consisting entirely of eigenvectors for $\mathbf A$.


Tensors are generalizations of scalars, vectors, and matrices. A tensor is distinguished by its rank, which is the number of indices it has. A scalar is a $0^{th}$-rank tensor (it has no indices), a vector is a $1^{th}$-rank tensor, i.e. its components are accessed by one index, e.g. $x_i$, and a matrix is a $2^{th}$-rank tensor, i.e. its components are accessed by two indices, e.g. $X_{i,j}$, and so on.

Just as we have scalar and vector fields, we also have tensor fields.