NoteNextra-origin/content/Math429/Math429_L21.md

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)