Lecture 27 - Finishing Minimal Polynomials & Starting Jordon Block Decomposition

Recall that $p_{T} (z)$ is the characteristic polynomial of $T$ while $μ_{T} (z)$ is the characteristic polynomial of $T$ .

We derived some facts:

Fact

$\deg (μ_{T}) \leq \deg (p_{T}) = \dim (V)$
Any polynomial $s$ with $s (T) = {\vec{0}}_{T}$ can only happen iff $s$ is a multiple of $μ_{T}$ .
$p_{T}$ is a multiple of $μ_{T}$
The zeroes of $μ_{T}$ are precisely the eigenvalues of $T$ .

The proof is as follows:

Proof
(1): From the definition.

(2): $\Leftarrow$ is trivial. For $\Rightarrow$ suppose $s$ such that $s (T) = \vec{0}$ . Write:

s = p μ_{T} + r

using the Division algorithm for polynomials. Plug in $T$ on both sides:

s (T) = p (T) μ_{T} (T) + r (T) = p (T) \vec{0} + r (T) = r (T) = \vec{0}

Thus $r (T) = \vec{0}$ so then $s$ is a multiple of $μ_{T}$ .

(3): Apply (2)

(4): Suppose $λ$ is a zero of $μ_{T} (z) = z^{m} + α_{m - 1} z^{m - 1} + \dots + α_{1} z + α_{0}$ . Thus $z - λ$ is a factor of $μ_{T}$ so then:

μ_{T} (z) = (z - λ) q (z), \deg (q) = \deg (μ_{T}) - 1

Plugging in $T$ :

q (T) \neq \vec{0}

since $μ_{T}$ is a minimal polynomial. But then:

μ_{T} (T) = (T - λ I) q (T) = \vec{0}

Thus $q (T) \neq \vec{0}$ , so there's some vector $v$ such that $q (T) v \neq \vec{0}$ . As such, then:

\vec{0} = (T - λ I) \underset{\neq \vec{0}}{\underset{⏟}{q (T) v}}

So $λ$ is an eigenvalue of $v$ via $T$ . This shows $\Rightarrow$ . For $\Leftarrow$ suppose $λ$ is an eigenvalue. Then $T v = λ v$ for some $v \neq \vec{0}$ . Then:

μ_{T} (T) = T^{m} - α_{m - 1} T^{m - 1} + \dots + α_{0} I

Plug in $v$ :

\begin{aligned} \vec{0} & = λ^{m} v + α_{m - 1} λ^{m - 1} v + \dots + α_{0} v \\ = (λ^{m} + α_{m - 1} λ^{m - 1} + \dots + α_{0}) \underset{\neq \vec{0}}{\underset{⏟}{v}} \end{aligned}

Thus $μ_{T} (λ) = 0$ .
☐
This implies that if you have the characteristic polynomial $p_{T}$ , then you kinda know what the minimal polynomial $μ_{T}$ should look like. Suppose:

p_{T} (z) = \prod_{i = 1}^{m} (z - λ_{i})^{d_{i}}

Then $μ_{T} (z)$ is:

μ_{T} (z) = \prod_{i = 1}^{m} (z - λ_{i})^{c_{i}}

where $1 \leq c_{i} \leq d_{i}$ for all $i$ .

Example

Suppose $T \in L (R^{3})$ where:

M (T) = [\begin{matrix} 5 & 1 & - 2 \\ - 39 & - 5 & 23 \\ 3 & 1 & 0 \end{matrix}]

Let $v_{1} = (1, - 4, 1)$ , $v_{2} = (0, - 1, 0)$ and $v_{3} = (2, - 1, 3)$ compose the basis $β$ . Then:

M (T, β) = [\begin{matrix} - 1 & 1 & - 1 \\ 0 & - 1 & - 6 \\ 0 & 0 & 2 \end{matrix}]

Thi is an UT matrix so $λ_{1} = - 1$ with $d_{1} = 2$ and $λ_{2} = 2$ with $d_{2} = 1$ . We could further find the basis $β^{'}$ such that we have block-diagonal matrices. However, here:

p_{T} (z) = (z + 1)^{2} (z - 2)

But:

μ_{T} (z) = (z + 1)^{α} (z - 2)

where $α \in {1, 2}$ . But which one? We calculate for the $α = 1$ case:

(T + I) (T - 2 I) = T^{2} - T - 2 I

We could find the matrix representation and find that $T^{2} - T - 2 I \neq \vec{0}$ , so then $(+ 1) (z - 2)$ is not the minimal polynomial. Thus, the characteristic polynomial in this case is the minimal polynomial.

8.D: Jordan Decomposition

While Axler doesn't use a proof that uses quotient spaces, we'll use them here since we are good at them.

Recall that for any operator $T \in L (V)$ that there is some $k$ such that:

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

where if $T$ is nilpotent then:

null (T^{k}) = \dots = V

We know that we can find a basis $β$ such that $M (T, β)$ is "extra upper triangular" where we have upper triangular sub-matrices in the matrix.

The question here is: can we do better? Namely, can we get more zeroes in $M (T, β)$ ?

Goal

We want to produce a basis such that $M (T, β)$ has many zeros.

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.