NoteNextra-origin/content/Math429/Math429_L22.md

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

Tv_k=((T-\lambda_m I)+\lambda_m I)v_k=(T-\lambda_m I)v_k+\lambda_m v_k

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.

M(T)=\begin{pmatrix}
    M(T\vert_U) &\vert & *\\
    \rule{2cm}{1pt}&&\rule{4cm}{0.4pt}\\
    0  & \vert&\lambda \textup{ on the diagonal line}
\end{pmatrix}

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

E(\lambda, T)=null\ (T-\lambda I)=\{ v\in V\vert Tv=\lambda v\}

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

E(\lambda_1, T)+...+E(\lambda_m,T)

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

dim\ (E(\lambda_1, T))+...+dim\ (E(\lambda_m,T))\leq dim\ V

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)