Lecture 28 - Continuing Jordan Block Decomposition

Last time we looked at the following:

An Example

Let's do an example to get our feet wet. Suppose $T \in L (V)$ with $\dim (V) = 9$ . Further, we suppose that: $\underset{\dim = 0}{\underset{⏟}{null (T^{0})}} \subset \underset{\dim = 4}{\underset{⏟}{null (T^{1})}} \subset \underset{\dim = 7}{\underset{⏟}{null (T^{2})}} \subset \underset{\dim = 8}{\underset{⏟}{null (T^{3})}} \subset \underset{\dim = 9}{\underset{⏟}{null (T^{4})}} = \dots = V$
So since $T^{0}, T^{1}, T^{2}, T^{3}$ have non-trivial nullspaces, we can choose vectors from each of them:

Choose $v_{1} \in V$ such that $T^{3} v \neq \vec{0}$ . Note that $T^{4} v_{1} = \vec{0}$
Now consider $null (T^{4}) / null (T^{3})$ . They are vectors which are in $null (T^{4})$ plus something in $null (T^{3})$ . But we know that $v_{1} \notin null (T^{3})$ so then $v_{1} + null (T^{3}) \neq \vec{0} + null (T^{3})$ is a valid basis for it.
Now consider $null (T^{3}) / null (T^{2})$ , $T v_{1} + null (T^{2})$ is a basis for $null (T^{3}) / null (T^{2})$ .
Consider $null (T^{2}) / null (T)$ , then $T^{2} v_{1} + null (T)$ is a start, but we need to add 2 more vectors. Thus add $v_{2}, v_{3}$ in this list. So we have $T^{2} v_{1} + null (T), v_{2} + null (T), v_{3} + null (T)$ .
Consider $null / {\vec{0}}$ . It's dimension is 4 so have $T^{3} v_{1}, T v_{2}, T v_{3}, v_{4}$ where $v_{4}$ is added on.

In general:

Choose $v_{1} \in V$ such that $T^{n - 1} v \neq \vec{0}$ for $n = \dim (V)$ .
Find a basis for $null (T^{n - k}) / null (T^{n - k - 1})$ where $k \geq 1$ .
Repeat for growing values of $k$ , moving over the difference in dimensions of our nullspace chain.

Notice that the dimension of the first quotient space was 1, then 1, then 3, then 4, taking the differences of each dimension as we start from the right and move to the left. In fact, we did a HW Problem that showed a similar part of this relationship, showing the iterative step:

13

Theorem

Suppose $U$ is a subspace of $V$ and $v_{1} + U, . . ., v_{m} + U$ is a basis of $V / U$ and $u_{1}, . . ., u_{n}$ is a basis of $U$ . Then $v_{1}, . . ., v_{m}, u_{1}, . . ., u_{n}$ is a basis of $V$ .

Proof
Let's determine if it's a basis by equating dimension, and determining LI.

First, the dimension of $V / U$ is $m$ given by the first basis, and the dimension of $U$ is $n$ via the second basis. As such, then:
$dim (V / U) = dim (V) - dim (U) \Rightarrow dim (V) = m + n$
so the proposed basis has the right number of vectors from $V$ .

Now for LI. Consider when:
$α_{1} v_{1} + \dots + α_{m} v_{m} + α_{m + 1} u_{1} + \dots + α_{m + n} u_{n} = \vec{0}$
Notice that our first basis for $V / U$ implies that for any $a_{i} \in F$ that if:
$a_{1} (v_{1} + U) + \dots + a_{m} (v_{m} + U) = ((a_{1} v_{1}) + U) + \dots + ((a_{m} v_{m}) + U) = \vec{0} = U$
Then all $a_{i} = 0$ , which becomes:
$(a_{1} v_{1} + \dots + a_{m} v_{m}) + U = U \Rightarrow a_{1} v_{1} + \dots + a_{m} v_{m} = \vec{0}$
So namely we've found that $v_{1}, . . ., v_{m}$ must be LI. We already know that the same thing occurs for $u_{1}, . . ., u_{n}$ showing those vectors are LI.

Notice that if $α_{m + 1} u_{1} + \dots + α_{m + n} u_{n} = \vec{0}$ then we are done since then:
$α_{1} v_{1} + \dots + α_{m} v_{m} = \vec{0}$
So all $α_{i} = 0$ as required. Now instead, say that $α_{m + 1} u_{1} + \dots + α_{m + n} u_{n} \neq \vec{0}$ , call it $u \in U$ . We'll show that this is impossible via contradiction. So then:
$α_{1} v_{1} + \dots + α_{m} v_{m} + u = \vec{0}$
Clearly if $α_{1} v_{1} + \dots + α_{m} v_{m} = \vec{0}$ then we'd have $u = \vec{0}$ which is a contradiction, so suppose the sum equals $v$ . Then:
$v + u = \vec{0} \Rightarrow v = - u$
But this is a contradiction! This is because then $v \in U$ since $v = - u$ . That implies that $v + U = U$ , which implies that $\vec{0} = v$ which is a contradiction. As such, we must have one of $v, u = \vec{0}$ , so then we get that all $a_{i} = 0$ and thus the list of vectors are LI.

As such, the list of vectors is LI and of size $m + n$ , so then it's a valid basis for $V$ .
☐

In the end, we get the basis:

β = {\underset{cycle 1}{\underset{⏟}{T^{3} v_{1}, T^{2} v_{1}, T v_{1}, v_{1}}}, \underset{cycle 2}{\underset{⏟}{T v_{2}, v_{2}}}, \underset{cycle 3}{\underset{⏟}{T v_{3}, v_{3}}}, \underset{cycle 4}{\underset{⏟}{v_{4}}}}

This basis is actually called a Jordan-Basis. Notice that the first vector in each cycle is an eigenvector since, for example:

T (T^{3} v_{1}) = T^{4} v_{1} = \vec{0}

since $v_{1} \in null (T^{4})$ . We can repeat this process.

This will create a block diagonal matrix:

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

notice that the matrix for $T - λ I$ would just have $- λ$ 's on the diagonal.

Proposition

Given a nilpotent operator $N$ $\exists v_{1}, . . ., v_{k}$ and corresponding non-negative integers $m_{1}, . . ., m_{k}$ such that:

$\underset{cycle 1}{\underset{⏟}{N^{m_{1}} v_{1}, N^{m_{1} - 1} v_{1}, . . ., v_{1}}}, N^{m_{2}} v_{2}, . . ., v_{2}, . . ., \underset{cycle k}{\underset{⏟}{N^{m_{k}} v_{k}, . . ., v_{k}}}$ is a basis for $V$
$N^{m_{i} + 1} v_{i} = \vec{0}$ for all $1 \leq i \leq k$ , so the $N^{m_{i}} v_{i}$ 's are eigenvectors of $λ = 0$ .

Jordan Form

Recall that $V = ⨁_{i = 1}^{m} G (λ_{i}, T)$ . If we happen to find bases $β_{1}, . . ., β_{m}$ where each $β_{i}$ is a basis for $G (λ_{i}, T)$ , then:

M (T, (β_{1}, . . ., β_{m}))

is block diagonal. And one block looks like $T |_{G (λ_{i}, T)} = (T - λ_{i} I) + λ_{i} I$ where $(T - λ_{i} I)$ is nilpotent while $λ_{i} I$ is diagonalizable.

Each block is called a Jordan-Block. To save notation we define:

J_{λ, n} = [\begin{matrix} λ & 1 & 0 & \dots & 0 \\ 0 & λ & 1 & \dots & 0 \\ 0 & 0 & λ & \dots & 0 \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & 0 & \dots & λ \end{matrix}]

where here $J_{λ, n}$ is $n \times n$ in size.

THEREFORE if you have any $T \in L (V)$ then we make the bases $β$ up above such that:

M (T) = [\begin{matrix} J_{λ_{1}, \dim G (λ_{1}, T)} & 0 & \dots & 0 \\ 0 & J_{λ_{2}, \dim G (λ_{2}, T)} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & J_{λ_{m}, \dim G (λ_{m}, T)} \end{matrix}]

And notice that each $J_{λ_{j}, n}$ is ALSO block diagonal.