NoteNextra-origin/pages/Math429/Math429_L36.md

Lecture 36

Chapter VIII Operators on complex vector spaces

Generalized Eigenvectors and Nilpotent Operators 8A

If T\in \mathscr{L}, is an linear operator on V and n=dim\ V.

\{0\}\subset null\ T\subset null\ T^2\subset \dots\subset null\ T^n=null\ T^{n+1}

Definition 8.14

T is called a nilpotent operator if null\ T^n=V. Equivalently, there exists k>0 such that T^k=0

Lemma 8.16

T is nilpotent \iff 0 is the only eigenvalue of T.

If \mathbb{F}=\mathbb{C}, then 0 is the only eigenvalue \implies T is nilpotent.

Proof:

If T is nilpotent, then T^k=0 for some k. The minimal polynomial of T is z^m=0 for some m. So 0 is the only eigenvalue.

over \mathbb{C}, the eigenvalues are all the roots of minimal polynomial.

Proposition 8.17

The following statements are equivalent:

1. T is nilpotent.
2. The minimal polynomial of T is z^m for some m\geq 1.
3. There is a basis of V such that the matrix of T is upper triangular with 0 on the diagonal (\begin{pmatrix}0&\dots&*\\ &\ddots& \\0 &\dots&0\end{pmatrix}).

Generalized Eigenspace Decomposition 8B

Let T\in \mathscr{L}(V) be an operator on V, and \lambda be an eigenvalue of T. We want to study T-\lambda I.

Definition 8.19

The generalized eigenspace G(\lambda, T)=\{(T-\lambda I)^k v=0\textup{ for some }k\geq 1\}

Lemma 8.20

G(\lambda, T)=null\ (T-\lambda I)^{dim\ V}

Proposition 8.22

If \mathbb{F}=\mathbb{C}, \lambda_1,...,\lambda_m all the eigenvalues of T\in \mathscr{L}, then

(a) G(\lambda_i, T) is invariant under T.
(b) (T-\lambda_1)\vert_{G(\lambda_1,T)} is nilpotent.
(c) V=G(\lambda_1,T)\oplus...\oplus G(\lambda_m,T)

Proof:

(a) follows from T commutes with T-\lambda_1 I. If (T-\lambda_1 I)^k=0, then (T-\lambda_i T)^k T(v)=T((T-\lambda_i T)^kv)=0

(b) follow from lemma

(c) V=G(\lambda_1,T)\oplus...\oplus G(\lambda_m,T)

1. V has a basis of generalized eigenvectors \implies V=G(\lambda_1,T)+...+G(\lambda_m,T)
2. If there exists v_i\in G(\lambda_i,T), and v_1+...+v_m=0, then v_i=0 for each i. Because the generalized eigenvectors from distinct eigenvalues are linearly independent, V=G(\lambda_1,T)\oplus...\oplus G(\lambda_m,T).

Definition 8.23

Let \lambda be an eigenvalue of T, the multiplicity of \lambda is defined as mul(x):= dim\ G(\lambda, T)=dim\ null\ (T-\lambda I)^{dim\ V}

Lemma 8.25

If \mathbb{F}=\mathbb{C},

\sum^n_{i=1} mul\ (\lambda_i)=dim\ V

Proof from proposition part (c).

Definition 8.26

If \mathbb{F}=\mathbb{C}, we defined the characteristic polynomial of T to be

q(z):=(z-\lambda_1)^{mul\ (\lambda_1)}\dots (z-\lambda_m)^{mul\ (\lambda_m)}

deg\ q=dim\ V, and roots of q are eigenvalue of V.

Theorem 8.29 Cayley-Hamilton Theorem

Suppose \mathbb{F}=\mathbb{C}, T\in \mathscr{L}(V), and q is the characteristic polynomial of T. Then q(T)=0.

Proof:

q(T)\in \mathscr{L}(V) is a linear operator. To show q(T)=0 it is enough to show q(T)v_1=0 for a basis v_1,...,v_n of V.

Since V is a sum of vectors in G(\lambda_1, T),...,G(\lambda_m,T).

q(T)=(T-\lambda_1 I)^{d_1}\dots (T-\lambda_m I)^{d_m}

The operators on the right side of the equation above all commute, so we can move the factor (T-\lambda_k I)^{d_k} to be the last term in the expression on the right. Because (T-\lambda_k I)^{d_k}\vert_{G(\lambda_k,T)}= 0, we have q(T)\vert_{G(\lambda_k,T)} = 0, as desired.

Theorem 8.30

Suppose \mathbb{F}=\mathbb{C}, T\in \mathscr{L}(V). Then the characteristic polynomial of T is a polynomial multiple of the minimal polynomial of T.