NoteNextra-origin/content/Math429/Math429_L23.md

Lecture 23

Chapter V Eigenvalue and Eigenvectors

5D Diagonalizable Operators

Theorem 5.55

Suppose V is a finite dimensional vector space and T\in \mathscr{L}(V). let \lambda_1,...,\lambda_m be the distinct eigenvalues of T, then the followings are equal:

a) T is diagonalizable b) V has a basis of eigenvectors of T c) V=E(\lambda, T)\oplus....\oplus E(\lambda_m,T) d) dim\ V= dim\ E(\lambda_1,T)+...+dim\ E(\lambda_m,T)

ideas of Proof:

(a)\iff (b) look at $M(T)$' (b)\iff (c) recall E(\lambda_1,T)+...+E(\lambda_m,T) is always a distinct sum (c)\iff (d) again E(\lambda_1,T)+...+E(\lambda_m,T) is always a distinct sum

Example: T:\mathbb{R}^2\to\mathbb{R}^3, $M(T)=\begin{pmatrix} 0&1&0\ 0&0&1\ 0&0&0 \end{pmatrix}$

Eigenvalues:[0], Eigenvectors E(0,T)=null\ (T-0I)=Span\{(1,0,0)\}

There are no basis of eigenvectors, \mathbb{R}^3\neq E(0,T), 3\neq dim\ (E(0,T))=1

Theorem 5.58

Suppose V is a finite dimensional T\in \mathscr{L}(v) and T has n=\dim . distinct eigenvalues then T is diagonalizable.

Proof:

Let \lambda_1,...,\lambda_n be the distinct elements of$T$.. =

Then let v_1,...,v_n be eigenvectors of \lambda_1,...,\lambda_n in the same order. Note v_1,...,v_n are eigenvectors for distinct eigenvectors by Theorem 5.11 they are linearly independent thus they form a basis. So by Theorem 5.55, T is diagonalizable.

Example:

M(T)=\begin{pmatrix}
1& 4& 5 \\
0&2&6\\
0&0&3
\end{pmatrix}  

is diagonalizable

Theorem 5.62

Suppose V finite dimensional T\in \mathscr{L}(V). Then T is diagonalizable if and only if the minimal polynomial is of the form (z-\lambda_1)...(z-\lambda_m) for distinct \lambda_1,...,\lambda_m\in\mathbb{F}

Proof:

\Rightarrow Suppose T is diagonalizable, let \lambda_1,...,\lambda_m be the distinct eigenvalues of T. And let v_1,...,v_n for n=dim\ V be a basis of eigenvectors of T. We need to show

(T-\lambda_1I)...(T-\lambda_mI)=0  

Consider (T-\lambda_1I)...(T-\lambda_mI)v_k=(T-\lambda_1I)...(T-\lambda_mI), suppose Tv_k=\lambda_j v_k. Then (T-\lambda_1I)...(T-\lambda_mI)=0

So (T-\lambda_1I)...(T-\lambda_mI)=0\implies minimal polynomial divides (z-\lambda_1)...(z-\lambda_m) so the minimal polynomial has distinct linear factors.

\Leftarrow

Suppose T has minimal polynomial (z-\lambda_1)...(z-\lambda_m) with distinct \lambda_1,...,\lambda_m

Induction on m,

Base case: (m=1):

Then T-\lambda I=0, so T=\lambda I is diagonalizable.

Induction step: (m>1):

Suppose the statement hold for <m, consider U=range\ (T-\lambda_mI), T\vert_U has minimal polynomial (z-\lambda_1)...(z-\lambda_m) so T\vert_U is diagonalizable.

null (T-\lambda_m I)=E(\lambda_m,T) has dim\ (E(\lambda_m,T)) distinct eigenvector.

Need to show null\ (T-\lambda_m I)\cap range\ (T-\lambda_m I)=\{0\}

Corollary

If U is an invariant subspace of T and T is diagonalizable, then T\vert_U is diagonalizable.

Proof:

minimal polynomial T\vert_U divides minimal polynomial of T.

Theorem (Gershigorem Disk Theorem)

The eigenvalue of T satisfies the following:

|\lambda-A_{j,j}|\leq \sum_{k=1,k\neq j}^n |A_{j_k}|

for some j where A=M(T)