Lecture 22 - Finishing G. Eigenspaces, Starting 8.B

Recall that we said that since $E (λ, T) = null (T - λ I)$ . This was an eigenspace corresponding to $λ$ . Similarly:

G (λ, T) = null (T - λ I)^{\dim (V)}

was the generalized eigenspace corresponding to $λ$ . Non-zero vectors $v \in G$ are called generalized eigenvectors.

Notice that if $λ$ is not an eigenvalue then we know that $null (T - λ I) = {\vec{0}}$ and thus we have it that $T - λ I$ is injective. Because injectivity composed on itself is injective then $(T - λ I)^{\dim (V)}$ must also be injective. Therefore:

λ is not an eigenvalue of T \Leftrightarrow G (λ, T) = {\vec{0}}

This suggests that even though after enough operations of $T$ we may add more eigenvectors, we cannot add eigenvalues. As such, there's really no such thing as "generalized eigenvalues" as if you called them that they're the exact same as just vanilla eigenvalues.

Similar to how different eigenvalue eigenvectors are LI, we have the following:

Proposition

Suppose $λ_{1}, . . ., λ_{m}$ are distinct eigenvalues of $T$ and $v_{1}, . . ., v_{m}$ are corresponding generalized eigenvectors. Then $v_{1}, . . ., v_{m}$ are LI.

Proof
Let $\dim (V) = n$ . Consider:

\sum_{i = 1}^{m} α_{i} v_{i} = \vec{0}

To show that $α_{i} = 0$ for all $i$ , let $S$ be the operator defined by:

S = \prod_{j \neq i} (T - λ_{j} I)^{n} = (T - λ_{1} I)^{n} \dots (T - λ_{i} I)^{n} \dots (T - λ_{m} I)^{n}

Applying $S$ to both sides, we can try to get rid of all the $α_{i} v_{i}$ 's except for one of them:

\begin{aligned} S (\sum_{i = 1}^{m} α_{i} v_{i}) & = \vec{0} \\ α_{i} S (v_{i}) & = \vec{0} \end{aligned}

We can explore the meaning of $S (v_{i})$ . Recall that $v_{i}$ is a generalized eigenvector corresponding to $λ_{i}$ . We can find the minimal $k_{i}$ such that:

(T - λ_{i} I)^{k_{i} + 1} v_{i} = \vec{0}

thus:

(T - λ_{i} I)^{k_{i}} v_{i} \neq \vec{0}

Call this left vector $w_{i}$ . Applying it:

(T - λ_{i} I) w_{i} = \vec{0} \Rightarrow w_{i} \in null (T - λ_{i} I)

So $w_{i}$ is an eigenvector of $T$ w/ $λ_{i}$ . Thus for any arbitrary $λ$ :

(T - λ I) w_{i} = T w_{i} - λ w_{i} = λ_{i} w_{i} - λ w_{i} = (λ_{i} - λ) w_{i}

So coming back to $S (v_{i})$ , apply $(T - λ_{i} I)^{k_{i}}$ to both sides:

(T - λ_{i})^{k_{i}} α_{i} S (v_{i}) = α_{i} S (w_{i}) = α_{i} \prod_{j \neq i} (λ_{i} - λ_{j})^{n} w_{i} = \vec{0}

Since all $j \neq i$ then $λ_{i} - λ_{j} \neq 0$ (remember, we have distinct eigenvalues, and $w_{i} \neq \vec{0}$ since it's an eigenvector, so then $α_{i} = 0$ for all $i$ . Thus we have LI.
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Interesting Characteristics of Nilpotent Operators

If $T$ is nilpotent, $\exists$ a basis of $V$ , denoted $β$ , for which $M (T, β)$ is an UT matrix with only zeroes on the diagonal.

Proof
Recall that we had:

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

For some $k \in Z^{+}$ . We know $null (T^{0}) = {\vec{0}}$ and we don't care. Consider the basis for $null (T)$ via $v_{1}^{'}, . . ., v_{β_{1}}^{'}$ . We can extend this to a basis of $null (T^{2})$ , and so on up to $null (T^{k})$ which has to be $V$ by the definition of being nilpotent (the range is the zero vector only, so then the nullspace is the whole space $V$ ):

\underset{basis for null (T^{k}) = V}{\underset{⏟}{\underset{basis for null (T^{2})}{\underset{⏟}{\underset{basis for null (T)}{\underset{⏟}{v_{1}^{'}, . . ., v_{β_{1}}^{'}}}, v_{1}^{2}, . . ., v_{β_{2}}^{2}}}, . . ., v_{1}^{k}, . . ., v_{β_{k}}^{k}}} = β

We can make a matrix for this:

We get the expected result because if you take any vector $v$ satisfying $T^{i} v = \vec{0}$ then that means that $T^{i - 1} (T v) = \vec{0}$ . In other words, when you input the vector $v^{i}$ to $T$ means that because $v^{1} \in null (T)$ then $T v^{1} = \vec{0}$ . For $v^{2} \in null (T^{2})$ then $T^{2} v^{2} = \vec{0}$ hence the diagonals being 0.
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As an example, we had the matrix yesterday:

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

Let $v = (1, 1, 1, 1)$ . $T v = (3, 2, 1, 0)$ . $T^{2} v = (3, 1, 0, 0)$ . $T^{3} v = (1, 0, 0, 0)$ . $T^{4} v = \vec{0}$ .

These 4 non-zero vectors $T^{3} v, T^{2} v, T v, v$ are all LI (the order here is just for the basis order for the next step). Since it's the same amount as the dimension of $R^{4}$ , then it's a basis. So even though $M (T)$ was with respect to the standard basis, notice that if we call this new basis $β^{'}$ :

M (T, β^{'}) = [\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}]

The cool thing is that we can get 0's above the "diagonal" of 1's we have. But this process is generalizable. If $T^{k} v \neq \vec{0}$ but $T^{k + 1} v = \vec{0}$ (ie: we have nilpotency) then the list $T^{k} v, . . ., T v, v$ is already guarunteed to be LI, and thus they are already a basis and we can make the matrix above for $T$ .

Where does the LI come from? Consider:

\begin{aligned} α_{k} T^{k} v + \dots + α_{1} T v + α_{0} v & = \vec{0} \\ T^{k} (L H S) & = \vec{0} \\ α_{0} v & = \vec{0} \\ α_{0} & = 0 \end{aligned}

We can repeat this with $T^{k - 1}, . . ., T$ on the LHS and get that each $α_{i} = 0$ and thus we have LI.