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:

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 $Tv_k=\lambda_k v_k+...+\lambda_1 v_1$, $(T-\lambda_k I)\in Span\ (v_1,...,v_k)$