77 lines
3.0 KiB
Markdown
77 lines
3.0 KiB
Markdown
# Lecture 21
|
|
|
|
## Chapter V Eigenvalue and Eigenvectors
|
|
|
|
### Minimal polynomial 5B
|
|
|
|
#### Odd Dimensional Real Vector Spaces
|
|
|
|
#### Theorem 5.34
|
|
|
|
Let $V$ be an odd dimensional real vector space and $T\in \mathscr{L}(V)$ a linear operator then $T$ has an eigenvalue.
|
|
|
|
#### Theorem 5.33
|
|
|
|
Let $\mathbb{F}=\mathbb{R}$, $V$ be a finite dimensional vector space. $T\in\mathscr{L}(V)$ then $dim\ null\ (T^2+bT+cI)$ is even for $b^2\leq 4c$.
|
|
|
|
Proof:
|
|
|
|
$null\ (T^2+bT+cI)$ is invariant under $T$, so it suffices to consider $V=null\ (T^2+bT+cI)$. Thus $T^2+bT+cI=0$.
|
|
|
|
Suppose $\lambda \in \mathbb{R}$ and $v\in V$ such that $Tv=\lambda v$, then if $v\neq 0$, then $z-\lambda$ must divide $z^2+bz+c$. but $z^2+bz+c$ does not factor over $\mathbb{R}$. Then we don't have eigenvalues.
|
|
|
|
Let $U$ be the **largest invariant subspace** of even dimension. Suppose $w\in V$ and $w\cancel{\in} U$ consider $W=Span\ (w,Tw)$ note $dim\ (w)=2$. Consider $dim(U+W)=dim U+dim W-dim(U\cap W)$.
|
|
|
|
So if $dim(U\cap W)=2$ then $w\in U$, which is a contradiction ($w\cancel{\in} U$).
|
|
|
|
If $dim(U\cap W)=1$ then $U\cap W$ invariant and gives an eigenvalue, which is a contradiction (don't have eigenvalues).
|
|
|
|
If $dim(U\cap W)=0$ $U+W$ is a larger even dimensional invariant subspace, which is a contradiction ($U$ be the **largest invariant subspace** of even dimension).
|
|
|
|
So $U=V$, $dim\ V$ is even.
|
|
|
|
### Upper Triangular Matrices 5C
|
|
|
|
#### Definition 5.38
|
|
|
|
A square matrix is **upper triangular** if all entries below the diagonal are zero.
|
|
|
|
Example:
|
|
|
|
$$
|
|
\begin{pmatrix}
|
|
1& 2& 3\\
|
|
0& 3 &4\\
|
|
0& 0& 5
|
|
\end{pmatrix}
|
|
$$
|
|
|
|
#### Theorem 5.39
|
|
|
|
Suppose $T\in \mathscr{L}(V)$ and $v_1,...,v_n$ is a basis, then the following are equal:
|
|
|
|
a) $M(T,(v_1,...,v_n))$ is upper triangular
|
|
b) $Span\ (v_1,...,v_n)$ is invariant $\forall k=1,...,n$
|
|
c) $Tv_k\in Span\ (v_1,...,v_n)$ $\forall k=1,...,n$
|
|
|
|
Sketch of Proof:
|
|
|
|
a)$\implies$c) is clear... (probably) b)$\iff$ c), then do c)$\implies$a), go step by step and construct $M(T,(v_1,...,v_n))$.
|
|
|
|
#### Theorem 5.41
|
|
|
|
Suppose $T\in\mathscr{L}(V)$ if there exists a basis where $M(T)$ is upper triangular with diagonal entries $\lambda_1,...,\lambda_n$, and $(T-\lambda _1 I)(T-\lambda_2 I)...(T-\lambda_n I)=0$, then $\lambda_1,...,\lambda_n$ are precisely the eigenvalues.
|
|
|
|
Proof:
|
|
|
|
Note that for $(T-\lambda_1 I)v_1=0$, consider $(T-\lambda_k I)v_k\in Span\ (v_1,...,v_{k-1})$, consider $w=Span\ (v_1,...,v_k)$ then $(T-\lambda_k I)\vert_w$ is not injective since $range\ (T-\lambda_k I)\vert_w=Span\ (v_1,...,v_{k-1})$, so $\lambda_k$ is an eigenvalue.
|
|
|
|
but the minimal polynomial divides $(z-\lambda_1)...(z-\lambda_n)$, so every eigenvalue is in.
|
|
|
|
#### Theorem 5.40
|
|
|
|
Suppose $T\in\mathscr{L}(V)$ if there exists a basis where $M(T)$ is upper triangular with diagonal entries $\lambda_1,...,\lambda_n$, then $(T-\lambda _1 I)(T-\lambda_2 I)...(T-\lambda_n I)=0$.
|
|
|
|
Proof:
|
|
|
|
Note that for $(T-\lambda_1 I)v_1=0$ and $Tv_k\in Span\ (v_1,...,v_k)$, and $Tv_k=\lambda_k v_k+...+\lambda_1 v_1$, $(T-\lambda_k I)\in Span\ (v_1,...,v_k)$ |