Lecture 23 - Decomposition of an Operator

The main result of this section is as follows.

Theorem

Given an operator $T \in L (V)$ where $V$ is a complex finite-dimensional vector space, then there is a basis of generalized eigenvectors of $T$ .

This is big! It says that any operator can be turned into a diagonal-matrix as a direct corollary:

Corollary

$V$ is a complex finite-dimensional vector space. Let $λ_{1}, . . ., λ_{m}$ be the distinct eigenvalues of $T \in L (V)$ . Then:

$V = G (λ_{1}, T) \oplus G (λ_{2}, T) \oplus \dots \oplus G (λ_{m}, T)$
Each $G (λ_{i}, T)$ is $T$ -invariant.
Each $(T - λ_{i} I) |_{G (λ_{i}, T)}$ is nilpotent.

In trying to prove this, we'll have to use one of these facts:

Fact

The null space and the range of a polynomial operator in $T$ are both $T$ -invariant.

We first prove the Lecture 23 - Decomposition of an Operator#^3c4724 proof:
\begin{proof}
(a) By induction on $\dim (V)$ .

The base case is $\dim (V) = 1$ . Then every non-zero vector is an eigenvector, so every basis chosen is a generalized eigenbasis.

For the inductive case, suppose $\dim (V) = n + 1$ and that the theorem holds for all $\dim (V) = n$ . Reminder that $T$ has an eigenvalue, because every operator on a complex space has an eigenvalue. Call it $λ_{1}$ .

Recall from yesterday:

Proposition

$V = null (T^{\dim (V)}) \oplus range (T^{\dim (V)})$

Thus:

V = null ((T - λ_{1} I)^{\dim (V)}) \oplus range ((T - λ_{1} I)^{\dim (V)})

The left part $null ((T - λ_{1} I)^{\dim (V)}) = G (λ_{1}, T)$ and let $U = range ((T - λ_{1} I)^{\dim (V)})$ . Notice that:

$U$ is invariant
$dim (U) < \dim (V)$
$λ_{1}$ is not an eigenvalues of $T |_{U}$ because the eigenvalues of $T |_{U}$ are $λ_{2}, . . ., λ_{m}$

Using the inductive hypothesis on the vector space $U$ and operator $T |_{U}$ :

U = G (λ_{2}, T |_{U}) \oplus \dots \oplus G (λ_{m}, T |_{U})

If we can show that $G (λ_{i}, T |_{U}) = G (λ_{i}, T)$ for all $2 \leq i \leq m$ then we are done.

We know $\subseteq$ because if you're in the left set, you're in $U$ so applying $T$ to those items is equivalent to $T |_{U}$ and thus we are done.

For $\supseteq$ , suppose $v = v_{n 1} + v_{r}$ where $v_{n 1} \in G (λ_{1}, T)$ and $v_{r} \in U$ . Because $U$ has it's own decomposition. Also (from inductive hypothesis):

v_{r} = v_{n 2} + \dots + v_{n m}

where $v_{n i} \in G (λ_{i}, T |_{U})$ and thus $v_{n i} \in G (λ_{i}, T)$ (the $\subseteq$ case). We have:

v = v_{n 1} + v_{n 2} + \dots + v_{n m}

where now each $v_{n i} \in G (λ_{i}, T)$ (notice the change from $T |_{U}$ ). All the $v_{n i}$ 's are linearly independent since they are all eigenvectors of different $λ_{i}$ values. Therefore, $v_{n 1}, . . ., v_{n m}$ is LI.

Thus for all $j \neq i$ we have $v_{n i} = v$ and $v_{n j} = \vec{0}$ . This suggests $v \in G (λ_{i}, T |_{U})$ and thus we are done.

(b) Using the definition of $G$ :

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

Where the right side is a polynomial operator $p (T)$ , so then it's $T$ -invariant.

(T - λ_{i} I) |_{G (λ_{i}, T)} = \vec{0} \leftarrow the zero operator

showing nilpotency. Notice that it doesn't mean that $T - λ_{i} I$ is nilpotent. The restriction is important.
\end{proof}

multiplicity of an eigenvalue

For $T \in L (V)$ , the multiplicity of an eigenvalue is $\dim (G (λ, T)) = \dim (null (T - λ I)^{\dim (V)})$ .

Note

We've talked about algebraic and geometric multiplicity of an eigenvalue in Linear I. Here the definition above is for the algebraic version. For the geometric one, it's defined as:

\dim (E (λ, T))

Fact

$\sum_{i = 1}^{k} G (λ_{i}, T) = \dim (V)$

The proof from this comes from (a) of the main theorem we looked at, using the fact that direct sums add up in dimension.