# Invariant subspaces, Eigenvalues, Eigenvectors

Aug-Nov 2020

## Recap

• Vector space $V$ over a scalar field $F$
• $F$: real field $\mathbb{R}$ or complex field $\mathbb{C}$ in this course
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• dim null $T$ + dim range $T$ = dim $V$
• Linear equation: $Ax=b$
• Solution (if it exists): $u+$ null$(A)$
• Four fundamental subspaces of a matrix
• Column space, row space, null space, left null space
• Determinant of a square matrix
• Function with many interesting properties
• Change of basis for linear map
• Can result in simpler matrix representation

## Invariant subspace

$T:V\to V$ is an operator

A subspace $U\subseteq V$ is said to be invariant under $T$ if $Tu\in U$ for all $u\in U$.

Examples

$A=\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}$

$\mathbb{R}^2$: invariant

$A=\begin{bmatrix} 2&0\\ 0&3 \end{bmatrix}$

span$\{(1,0)\}$: invariant, span$\{(0,1)\}$: invariant

$A=\begin{bmatrix} 1&2\\ -1&4 \end{bmatrix}$

Do invariant subspaces exist?

## Why invariant subspaces?

$T\leftrightarrow\begin{bmatrix} 1&2\\ -1&4 \end{bmatrix}$

span$\{(2,1)\}$: invariant, span$\{(1,1)\}$: invariant

Change of basis

$B=\{(2,1),(1,1)\}$

$T\leftrightarrow\begin{bmatrix} 2&0\\ 0&3 \end{bmatrix}$

Found a basis under which $T$ is a diagonal matrix

## Eigenvalues

one-dimensional invariant subspace

$T:V\to V$ is an operator, dim $V=n$, $T\leftrightarrow A$ in some basis

span$\{v\}$ is a one-dimensional invariant subspace iff $Tv=\lambda v$

$\lambda\in F$ is an eigenvalue of $T$ if there exists a $v$ such that $v\ne 0$ and $Tv=\lambda v$.

Do eigenvalues exist?

$Tv=\lambda v$, $v\ne0$ iff $(T-\lambda I)v=0$, $v\ne0$ iff $T-\lambda I$: non-invertible

$T-\lambda I$ non-invertible iff det$(A-\lambda I)$ = 0

det$(A-\lambda I)$: polynomial of degree $n$ in $\lambda$

Roots of det$(A-\lambda I)$ are eigenvalues of $T$.

Eigenvalues are for the operator $T$ and are the same under any basis

det$(S^{-1}AS-\lambda I)=$ det$(S^{-1}(A-\lambda I)S)=$ det$(A-\lambda I)$

## Eigenvectors

$T:V\to V$ is an operator

$v\in V$ is an eigenvector corresponding to an eigenvalue $\lambda\in F$ of $T$ if $v\ne 0$ and $Tv=\lambda v$.

• If $v$ is an eigenvector, then so is $\alpha v$ for $\alpha\ne 0$. The scaled versions are not considered new.

• For the same eigenvalue $\lambda$, linearly independent eigenvectors, if they exist, need to be found.

Eigenvectors of $\lambda$: basis of null $T-\lambda I$

Example 1

$A=\begin{bmatrix} 2&0\\ 0&3 \end{bmatrix}$

det$(A-\lambda I)=(2-\lambda)(3-\lambda)=0$

Eigenvalues: $2$ and $3$

Eigenvectors: for $2$, $(1,0)$; for $3$, $(0,1)$

## Examples and multiplicities

$A=\begin{bmatrix} 1&2\\ -1&4 \end{bmatrix}$

det$(A-\lambda I)=(1-\lambda)(4-\lambda)+2=\lambda^2-5\lambda+6=0$

Eigenvalues: $2$ and $3$

Eigenvectors: for $2$, $(2,1)$; for $3$, $(1,1)$

$A=\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}$

det$(A-\lambda I)=(1-\lambda)^2=0$

Eigenvalues: $1$ repeated twice

Eigenvectors: for $1$, $(1,0)$ and $(0,1)$

algebraic multiplicity of eigenvalue $\lambda$: mutliplicity of root $\lambda$ in det$(A-\lambda I)$

geometric multiplicity of eigenvalue $\lambda$: dim null $T-\lambda I$

## Examples

$A=\begin{bmatrix} 1&2\\ 0&1 \end{bmatrix}$

det$(A-\lambda I)=(1-\lambda)^2=0$

Eigenvalues: $1$ repeated twice

Eigenvectors: for $1$, $(1,0)$

Algebraic multiplicity = 2 and Geometric multiplicity = 1

General case: $n\times n$ matrix $A$

1. Modern numerical methods can compute eigenvalues and eigenvectors numerically.

2. In most cases, det$(A-\lambda I)$ is not a preferred method.

3. Determinant approach is useful for small cases and hand calculations in quizzes!

## Linear independence of eigenvectors

Let $v_1,\ldots,v_m$ be eigenvectors of distinct eigenvalues $\lambda_1,\ldots,\lambda_m$. Then, $v_1,\ldots,v_m$ are linearly independent.

Proof

Let $k$ be the least value such that $v_k\in$ span$\{v_1,\ldots,v_{k-1}\}$ $v_k=a_1v_1+\cdots+a_{k-1}v_{k-1}$

Applying $T-\lambda_k I$, $0=a_1(\lambda_1-\lambda_k)v_1+\cdots+a_{k-1}(\lambda_{k-1}-\lambda_k)v_{k-1}$

Contradicts with $k$ being least value chosen above