# 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 $A=M(T)$