Basics of Linear Algebra

Linearity is the simplest structure in mathematics. Let's review the basic notations and terminologies in linear algebra.

$A$ is an $m\times n$ real matrix, written $A\in {\mathbb{R}}^{m\times n}$, if

$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& & & ⋮\\ a_{m1}& a_{m2}& \cdots & a_{mn}\end{array}\right]$$

where $a_{ij}\in \mathbb{R}$. The $(i,j)$th entry of $A$ is $A_{ij}=a_{ij}$.

The transpose of $A\in {\mathbb{R}}^{m\times n}$ is defined as

$$A^T=\left[\begin{array}{cccc}{A}_{11}& A_{21}& \cdots & A_{m1}\\ A_{12}& A_{22}& \cdots & A_{m2}\\ ⋮& & & ⋮\\ A_{1n}& A_{2n}& \cdots & A_{mn}\end{array}\right]\in {\mathbb{R}}^{n\times m}$$

In other words, $(A^T{)}_{ij}=A_{ji}$.

Note: $\mathbf{x}\in {\mathbb{R}}^n$ is considered to be a column vector in ${\mathbb{R}}^{n\times 1}$.

Definition(Sums and products of matrices). The sum of matrices $A\in {\mathbb{R}}^{m\times n}$ and $B\in {\mathbb{R}}^{m\times n}$ is the matrix $A+B\in {\mathbb{R}}^{m\times n}$ such that

$$(A+B{)}_{ij}=A_{ij}+B_{ij}$$

The product of matrices $A\in {\mathbb{R}}^{m\times n}$ and $B\in {\mathbb{R}}^{n\times \ell }$ is the matrix $AB\in {\mathbb{R}}^{m\times \ell }$ such that

$$(AB{)}_{ij}=\sum _{k=1}^n{A}_{ik}{B}_{kj}$$

The $n\times n$ identity matrix, denoted $I_n$ or $I$ for short, is

$$I=I_n=\left[\begin{array}{cccc}1& 0& \cdots & 0\\ 0& 1& \cdots & 0\\ ⋮& & & ⋮\\ 0& 0& \cdots & 1\end{array}\right]\in {\mathbb{R}}^{n\times n}$$

We have $IA=A=AI$.

If it exists, the inverse of $A$, denoted $A^{-1}$, is a matrix such that $A^{-1}A=I$ and $A{A}^{-1}=I$. If $A^{-1}$ exists, we say that $A$ is invertible. We have $(A^{-1}{)}^T=(A^T{)}^{-1}$ and $(AB{)}^{-1}=B^{-1}{A}^{-1}$.

The trace of a square matrix $A\in {\mathbb{R}}^{n\times n}$, denoted $\text{tr}(A)$, is defined as $\text{tr}(A)=\sum _{i=1}^n{A}_{ii}$. We have $\text{tr}(AB)=\text{tr}(BA)$ if $AB$ is a square matrix.

We say $A$ is symmetric if $A=A^T$.

The rank of a matrix $A$ is the dimension of $A$'s column space.

Numpy Matrix Operations

Numpy has optimized the matrix operations based on the BLAS and LAPACK libraries. So in python, for the most of the time, using Numpy to do matrix operations is faster than writing your own code.

import numpy as np

A = np.array([[1, 2], [3, 4]])
B = np.array([[5, 6], [7, 8]])

# Identity matrix
I = np.eye(2)
# Matrix addition
C = A + B
# Matrix multiplication
D = A @ B
# Matrix entry-wise multiplication
E = A * B 
# Matrix transpose
F = A.T
# Matrix inverse
G = np.linalg.inv(A)

Eigenvalues and Eigenvectors

Given two vectors $u,v\in {\mathbb{R}}^d$, the inner product is $⟨u,v⟩:=u^{T}v=\sum _{j=1}^d{u}_j{v}_j$. We define the ${\ell }_2$-norm of $u$ as $\Vert u\Vert =\sqrt{|⟨u,u⟩|}$. The cosine of the angle between $u$ and $v$ is $\cos \alpha =\frac{⟨u,v⟩}{\Vert u\Vert \Vert v\Vert }$. We say $u$ is orthogonal to $v$, denoted as $u\perp v$ if $⟨u,v⟩=0$.

Definition(Eigenvalues and Eigenvectors). We say $\lambda$ and $v$ are the eigenvalue and eigenvector of a matrix $A$ if $Av=\lambda v$.

Theorem(Eigenvalue Decomposition). A symmetric matrix $A\in {\mathbb{R}}^{d\times d}$ has:

• Real eigenvalues: ${\lambda }_1\ge \dots \ge {\lambda }_d$
• Orthonormal eigenvectors: $u_1,\dots ,u_d$ such that $\Vert u_i\Vert =1$ and $u_i\perp u_j$ for all $1\le i<j\le d$

such that we have the eigenvalue decomposition:

$$A=\sum _{i=1}^d{\lambda }_i{u}_i{u}_i^T$$

Denote $U=[u_1,\dots ,u_d]\in {\mathbb{R}}^{d\times d}$ and

$$\mathrm{\Lambda }=\text{diag}({\lambda }_1,\dots ,{\lambda }_d):=\left[\begin{array}{cccc}{\lambda }_1& 0& \cdots & 0\\ 0& {\lambda }_2& \cdots & 0\\ ⋮& & & ⋮\\ 0& 0& \cdots & {\lambda }_d\end{array}\right]$$

The eigenvalue decomposition can also be written as $A=U\mathrm{\Lambda }{U}^T$. We call $U$ an orthogonal matrix as $U^{T}U=I_d$.

As a cornerstone of our chapter, we prefer to interpret the concepts as a solution of an optimization problem. Such interpretation is usually called variational form. The following theorem gives a variational form of eigenvalues.

Theorem(Variational Form of Eigenvalues). Given a symmetric matrix $A$ same as above, its maximum eigenvalue has:

$${\lambda }_{max}(A):={\lambda }_1=\underset{\Vert x\Vert \le 1}{max}{x}^{T}Ax=\underset{x}{max}\frac{x^{T}Ax}{x^{T}x}$$

and $u_1={\text{argmax}}_{\Vert x\Vert \le 1}{x}^{T}Ax$

and its minimum eigenvalue has:

$${\lambda }_{min}(A):={\lambda }_d=\underset{\Vert x\Vert \le 1}{min}{x}^{T}Ax=\underset{x}{min}\frac{x^{T}Ax}{x^{T}x}$$

and $u_d={\text{argmin}}_{\Vert x\Vert \le 1}{x}^{T}Ax$

Question: What is the variational form of other eigenvalues? See the visualization in the figure above.

The concept of eigenvalue decomposition can be generalized to non-symmetric or even non-square matrices. We have the following theorem on singular value decomposition.

Theorem(Singular Value Decomposition). Given a rank $r$ matrix $A\in {\mathbb{R}}^{n\times d}$, there exist:

• Singular values: ${\sigma }_1\ge \dots \ge {\sigma }_r>0$ and denote $D=\text{diag}({\sigma }_1,\dots ,{\sigma }_r)$
• Orthogonal matrices: $U=[u_1,\dots ,u_r]\in {\mathbb{R}}^{n\times r}$ and $V=[v_1,\dots ,v_r]\in {\mathbb{R}}^{d\times r}$ satisfying $U^{T}U=V^{T}V=I_r$

such that we have the singular value decomposition (SVD):

$$A=UD{V}^T=\sum _{i=1}^r{\sigma }_i{u}_i{v}_i^T$$

We can see that the eigenvalue decomposition is a special SVD. The matrix $A$ as a linear map transforms the two canonical unit vectors (yellow and red vectors) in the top left plot to the top right plot. This map can be decomposed into three steps: 1. $V^T$: rotating yellow and red vectors to $v_1$ and $v_2$ 2. $D$: scaling the disc by ${\sigma }_1$ horizontally and ${\sigma }_2$ vertically 3. $U$: rotating two canonical unit vectors (blue lines) to $u_1$ and $u_2$

Given the concepts above, we can define the matrix operator norm.

Definition(Matrix Operator Norm). For a matrix $A\in {\mathbb{R}}^{n\times d}$, its operator norm (also called spectral norm) is defined as:

$$\Vert A{\Vert }_2=\underset{\Vert x\Vert \le 1}{max}\Vert Ax\Vert =\underset{\Vert x\Vert \le 1}{max}\sqrt{x^T{A}^{T}Ax}=\sqrt{{\lambda }_{max}(A^{T}A)}={\sigma }_1$$

where ${\sigma }_1$ is the largest singular value of $A$. The operator norm measures the maximum "stretching factor" of the matrix when applied to any unit vector. Geometrically, it represents the semi-major axis of the ellipse that the unit ball is mapped to under the linear transformation $A$. In the following, we will simply use $\Vert A\Vert$ to denote the operator norm of $A$.

Connection between SVD and Eigenvalue Decomposition:

• The singular values of $A$ are the square roots of the eigenvalues of $A^{T}A$: ${\sigma }_i(A)=\sqrt{{\lambda }_i(A^{T}A)}=\sqrt{{\lambda }_i(A{A}^T)}$.
• The left singular vectors of $A$ are the same as the eigenvectors of $A{A}^T$.
• The right singular vectors of $A$ are the same as the eigenvectors of $A^{T}A$.
• If $A$ is symmetric, it will have real eigenvalues and eigenvectors. Assume $|{\lambda }_1|\ge |{\lambda }_2|\ge \cdots \ge |{\lambda }_d|$, then ${\sigma }_i(A)=|{\lambda }_i|$. Note that we order the eigenvalues by their absolute values.
• If $A$ is positive semidefinite, it will have non-negative eigenvalues and eigenvectors and the SVD is the same as the eigenvalue decomposition (by ingoring the zero eigenvalues).