Lecture 22

Chapter V Eigenvalue and Eigenvectors

Upper Triangular Matrices 5C

Theorem 5.44

Let $T\in \mathscr{L}(V)$ be a linear operator, then $T$ has an upper triangular matrix (with respect to some basis), if the minimal polynomial is $(z-\lambda_1)...(z-\lambda_m)$ for $\lambda_1,..,\lambda_m\in \mathbb{F}$

Proof:

$\implies$ easy

$\impliedby$ Suppose the minimal polynomial of $T$ is $(z-\lambda_1)...(z-\lambda_m)$

Then we do induction on $m$.

Base case: $m=1$, then $T-\lambda_1 I=0$, $T=\lambda I$ but $\lambda I$ has an upper triangular matrix,

Induction step: $m>1$, Suppose the results holds for smaller $m$. Let $u=range(T-\lambda_m I)$, $U$ is invariant under $T$, consider $T\vert_u$.

Note that if $u\in U$, $(T-\lambda_1 I)...(T-\lambda_{m-1} I)u=(T-\lambda_1 I)...(T-\lambda_m I)v=0$. Thus the minimal polynomial of $T\vert_U$ divides $(z-\lambda_1)...(z-\lambda_{m-1})$

Corollary 5.47 (staring point for Jordan Canonical Form)

Suppose $V$ is a finite dimensional complex vector space, and $T\in \mathscr{L}(V)$, then $T$ has an upper triangular matrix with respect to some basis.

Recall: $T$ is upper triangular $\iff$ $Tv_k\in Span\ (v_1,...,v_k)$. where $v_1,...,v_n$ is a basis.

Let $u_1,...,u_r$ be a basis for $U$ such that $Tu_k\in Span\ (v_1,...,v_k)$ (such thing exists because $T$ is upper triangular.

Extend to a basis of $V$, $u_1,..,u_r,v_1,...,v_s$, then

and $(T-\lambda_m I)v_k\in U, \lambda_m v_k\in Span\ (u_1,..,u_r,v_k)$

Thus with respect to the same basis $u_1,..,u_r,v_1,...,v_s$ $T$ is upper triangular.

Example:

$M(T)=\begin{pmatrix} 2&0&1\\ 0&2&1\\ 1&1&3 \end{pmatrix}$ and the minimal polynomial is $(z-2)(z-2)(z-3)$

$v_1=(1,-1,0), v_2=(1,0,-1), v_3=(-1,1,0)$

$M(T,(v_1,v_2,v_3))=\begin{pmatrix} 2&1&0\\ 0&2&0\\ 0&0&3 \end{pmatrix}$ which is upper triangular.

5D Diagonalizable Operations

Definition 5.48

A Diagonal matrix is a matrix where all entries except the diagonal is zero

Example: $I,0,\begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&3 \end{pmatrix}$

Definition 5.50

An operator $T\in\mathscr{L}(V)$ is diagonalizable if $M(T)$ is diagonalizable with respect to some basis.

Example:

$T:\mathbb{F}->\mathbb{F^2}$

$M(T)=\begin{pmatrix} 3&-1\\ -1&3& \end{pmatrix} v_1=(1,-1), v_2=(1,1)$, $T(v_1)=(4,-4)=4v_1, T(v_2)=(2,2)=2v_2$, so the eigenvalues are $2$ with eigenvector $v_2$, and $4$ with eigenvector $v_1$. The eigenvectors for $z$ are $Span (v_2)\ \{0\}$

$M(T,(v_1,v_2))=\begin{pmatrix} 4&0\\ 0&2 \end{pmatrix}$ and $T$ is diagonalizable.

Definition 5.52

Let $T\in \mathscr{L}(V),\lambda \in \mathbb{F}$. the eigenspace of $T$ corresponding to $\lambda$ is the subspace $E(\lambda, T)\in V$ defined by

Example:

$E(2,T)=Span\ (v_2)$ $E(4,T)=Span\ (v_1)$, $E(3,T)=\{0 \}$

Theorem 5.54

Suppose $T\in \mathscr{L}(V)$ $\lambda_1,...,\lambda_m$ are distinct eigenvalues of $T$, Then

is a direct sum. In particular if $V$ is finite dimensional.

Proof:

Need to show that if $v_k\in E(\lambda_k,T)$ for $k=1,...,m$ then $v_1+...+v_m=0\iff v_k=0$ for $k=1,...,m$. i.e eigenvectors for distinct eigenvalues are linearly independent. (Prop 5.11)