More on Eigenvalues, Eigenvectors, Diagonalization

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
• Eigenvalue $\lambda$ and Eigenvector $v$: $Tv=\lambda v$
• Distinct eigenvalues have independent eigenvectors

Repeatedly applying an operator on a vector

$T:V\to V$, operator and $v\in V$, $v\ne 0$, $F=\mathbb{C}$ and dim $V=n$

$T$ can be applied repeatedly to $v$

$v$, $Tv$, $\ldots$, $T^nv$: $n+1$ vectors

Linearly dependent in $V$

$a_0v+a_1Tv+\cdots+a_nT^nv=0$, $a_i\in \mathbb{C}$

Let $m=$ max $i$ s.t. $a_i\ne 0$, $m\ge 1$

$(a_0+a_1T+\cdots+a_mT^m)v=0$, $v\ne0$, $a_m\ne 0$

Let $a_0+a_1x+\cdots+a_mx^m=a_m(x-\lambda_1)\cdots(x-\lambda_m)$

Operator algebra - they can be added, multiplied

$a_0+a_1T+\cdots+a_mT^m=a_m(T-\lambda_1I)\cdots(T-\lambda_mI)$

$a_m(T-\lambda_1I)\cdots(T-\lambda_mI)v=0$, $v\ne 0$

Existence of eigenvalues: another proof

Since $(T-\lambda_1I)\cdots(T-\lambda_mI)v=0$, $v\ne 0$, there exist at least one $i$ s.t. $T-\lambda_iI$ is non-invertible

Proof

Contradiction: if $T-\lambda_iI$ is invertible for every $i$, then $v=0$

$T-\lambda_iI$: non-invertible implies $\lambda_i$ is an eigenvalue

This proves existence of one eigenvalue for an operator ($F=\mathbb{C}$)

There are at most dim $V$ distinct eigenvalues for an operator

Proof

Eigenvectors of distinct eigenvalues are linearly independent

Eigenvalues of diagonal and triangular matrices

Diagonal elements of a diagonal matrix are eigenvalues of the corresponding operator

Proof

$T-\lambda I$: non-invertible if $\lambda=$ a diagonal element

Diagonal elements of a triangular matrix are eigenvalues of the corresponding operator

Proof

$T-\lambda I$: non-invertible if $\lambda=$ a diagonal element

Diagonalization

If $T$ has $n=$ dim $V$ distinct eigenvalues, the eigenvectors form a basis for $V$

Proof

Eigenvectors are independent for distinct eigenvalues

$n$ linearly independent eigenvectors form a basis

In the above basis of eigenvectors, $T$ is diagonal

Proof

$v_i$: eigenvector, $Tv_i=\lambda_i v_i$, $i=1,\ldots,n$

$\{v_1,\ldots,v_n\}$: basis

$Tv_i$: has $\lambda_i$ in $i$-th coordinate, zero elsewhere

Examples: Repeated Eigenvalues

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

Eigenvalues: $1, 1, 1$

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

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

Eigenvalues: $1, 1, 1$

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

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

Eigenvalues: $1, 1, 1$

Eigenvectors: $(1,0,0)$

Towards “simple” matrices

Eigenvalues are useful in obtaining simple matrices

Basis of eigenvectors results in a diagonal matrix

When eigenvalues are repeated, there may not be enough eigenvectors

What can be said, in general, about the “simplest” matrix for an operator?