Lecture 21 - Starting Generalized Eigenvectors and Nilpotent Operators

For the last couple of chapters, we've always worked in a:

Finite Dimensional
Inner Product Space

We will drop the inner product part and return to just pure vector spaces (still finite dimensional).

8.A: Generalized Eigenvectors and Nilpotent Operators

Context on Nilpotency

Starting with an example:

Example

Consider $T \in L (R^{4})$ , whose matrix representation w.r.t. the standard basis is:

M (T) = [\begin{matrix} 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}]

Then:

M (T^{2}) = M (T)^{2} = [\begin{matrix} 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]

Then:

M (T^{3}) = [\begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]

and finally $M (T^{4}) = \vec{0}$ and for higher powers

Such an operator such that there's some $k \in N$ where $T^{k} = \vec{0}$ implies that $T$ is nilpotent.

Generalized Eigenvectors

Given some $T \in L (V)$ , consider the sequence of subspaces:

null (T^{0}), null (T^{1}), \dots

Recall from HW 3 - Self-Adjoint and Normal Operators#7 that we can claim:

null (T^{i}) \subseteq null (T^{i + 1})

Because any $v$ on the LHS implies that $T^{i} v = 0 \Rightarrow T T^{i} v = 0$ so it's in the right side. As a result:

Finding 1

The sequence of subspaces is weakly increasing

Also notice that once' we start to get equality in a item compared to the next, then we must have that all items after are equal as well (since we get $\supseteq$ thus implying $=$ )

Finding 2

As soon as equality holds, then the sequence stabilizes.

Namely:

null (T^{0}) \subset null (T^{1}) \subset \dots \subset null (T^{i}) = null (T^{i + 1}) = \dots = \dots

Proof
Suppose $null (T^{j}) = null (T^{j + 1})$ . We want to show that $null (T^{j}) = null (T^{k})$ where $j < k$ .

Clearly $\subseteq$ is true since we can apply the first finding enough times to get this.

For $\supseteq$ , notice:

\begin{aligned} T^{k} v = \vec{0} & \Rightarrow \\ T^{j + 1} (T^{k - j - 1}) v = \vec{0} & \Rightarrow \\ T^{j} (T^{k - j - 1}) v = \vec{0} & \Rightarrow \\ \dots & \Rightarrow \\ T^{k - 1} v = \vec{0} & \Rightarrow \\ T^{k - 2} v = \vec{0} & \Rightarrow \\ \dots & \Rightarrow \\ T^{k - (k - j)} v = \vec{0} & \Rightarrow \\ T^{j} v = \vec{0} \end{aligned}

☐

Finding 3

The sequence must stabilize. Furthermore, it must stabilize by $null (T^{\dim (V)})$

Proof
Assume it didn't. We have:

null (T^{0}) \subset null (T^{1}) \subset \dots

But we have $dim (null (T^{i + 1})) > dim (null (T^{i}))$ . As such:

0 < dim (null (T)) < \dots < dim (null (T^{\dim (V)}))

But we can't keep going as the maximum dimension must be $\dim (V)$ .

At minimum, you always have to increase by 1, so at most you must stabilize once you apply $T$ $\dim (V)$ times.
☐

Bringing in the Nilpotents

An alternative way to describe nilpotency is that the sequence:

null (T^{0}), . . .

must stabalize to $V$ in order to be nilpotent (since once $T^{k} = \vec{0}$ is reached, then $null (T^{k}) = null (\vec{0}) = V$ ).

Fact

If $T$ is nilpotent, $T^{\dim (V)} = \vec{0}$ .

Proposition

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

Proof
First, we show the sum is direct. Suppose $v$ is in both sets, so $v \in null (T^{\dim (V)})$ and $v \in range (T^{\dim (V)})$ . From the former:

T^{\dim (V)} v = \vec{0}

and from the latter:

T^{\dim (V)} w = v

for some $w \in V$ . Plug the latter into the former:

T^{\dim (V)} (T^{\dim (V)} w) = \vec{0} \Rightarrow T^{2 \dim (V)} w = \vec{0} \Rightarrow T^{\dim (V)} w = \vec{0}

Because we can use our fact that the sequence stabalizes by $T^{\dim (V)}$ . This implies that $v = \vec{0}$ showing it as direct.

To show equality, just use dimensionality:

\text{dim}(\text{null}(T^{\dim(V)}) \oplus \text{range}(T^{\dim(V)})) = \text{dim}(\text{null}(T^{\dim(V)})) + \text{dim}(\text{range}(T^{\dim(V)})) = \text{dim}(V) {#db8500}

Using first the direct sum property, and then at the end the FTOLM.

If $T$ is nilpotent, the case is obvious since the left becomes $V$ and the right becomes ${\vec{0}}$ . If $T$ is not nilpotent,
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Clearly $null (T)$ is an important subspace of $T$ . We know that:

{\vec{0}} \subset null (T) \subset \dots null (T^{j}) = null (T^{j + 1}) = \dots

It seems that the first item $null (T)$ and $null (T^{j})$ are both important properties of $T$ . We can think of $null (T^{j})$ as all the vectors that eventually will get mapped to $\vec{0}$ through repeated applications of $T$ . Some just may not get sent to 0 right away.

Notice that $null (T)$ was the eigenspace of all $λ = 0$ :

null (T) = E (0, T)

but what about $null (T^{j})$ :

null (T^{j}) = G (0, T)

here $G$ is the name for the generalized eigenspace. Notice that since $null (T) \subseteq null (T^{j})$ then:

E (0, T) \subseteq G (0, T)

so let's officially define generalized eigenvectors/spaces:

generalized eigenvectors/spaces

A vector $v$ is a generalized eigenvector of $T$ where $T \in L (V)$ corresponding to $λ$ if
$v \neq \vec{0}$ and $(T - λ I)^{j} v = \vec{0}$ for some $j \in N$ .