Lecture 29 - Review of Jordan-Block, Complexification

Recall the Jordan form of a matrix. Namely $T \in L (V)$ then:

V = ⨁_{i = 1}^{m} G (λ_{i}, T)

where $λ_{1}, . . ., λ_{m}$ are the distinct eigenvalues of $T$ . Then $\exists$ a Jordan basis where:

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

where each $J_{λ, n}$ is:

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

The idea is that we consider each generalized eigenspace's basis $G (λ_{i}, T)$ and notice that $(T - λ_{i} I)$ is nilpotent and $λ_{i} I$ was diagonal:

T |_{G (λ_{i}, T)} = \underset{nilpotent on G (λ_{i}, T)}{\underset{⏟}{T - λ_{i} I}} + \underset{diagonal}{\underset{⏟}{λ_{i} I}}

This matrix is in Jordan Form, and any basis $β$ that produces this form is called a Jordan Basis. But notice that the numbers in the matrix are unique. But the order of the blocks and sub-blocks are inconsequential, so $M$ is not unique. Just choose a basis $β$ where the basis vectors that are blocked together are swapped around, and you'll see this effect.

Application: Higher Powers of Matrices

The nice thing about diagonalizability was that we could find higher powers of the transforamation really easily. We can do something similar for Jordon-Block matrices.

Example

Solve $a_{n} = 4 a_{n - 3} + 7 a_{n - 2} + 2 a_{n - 1}$ where $a_{0} = - 1, a_{1} = 2, a_{3} = 3$ . Namely, find an explicit formula for $a_{n}$ .

To solve this build a vector:

\vec{a_{n}} = [\begin{matrix} a_{n} \\ a_{n + 1} \\ a_{n + 2} \end{matrix}] \Rightarrow \vec{a_{n + 1}} = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 4 & 7 & 2 \end{matrix}] \vec{a_{n}} = [\begin{matrix} a_{n + 1} \\ a_{n + 2} \\ a_{n + 3} \end{matrix}]

where the bottom row comes from using the equation, solving for $a_{n + 3}$ . Call the matrix we made $A$ . In this case, $A$ is not-diagonalizable; however, notice that:

(A + I) [\begin{matrix} 2 \\ - 1 \\ 0 \end{matrix}] = [\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}]

(A + I) [\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

Thus $(- 2, 1, 0)$ and $(1, - 1, 1)$ are generalized eigenvectors with eigenvalue $λ = - 1$ in this case. Furthermore:

(A - 4 I) [\begin{matrix} 1 \\ 4 \\ 16 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

So $(1, 4, 16)$ is a generalized eigenvector with eigenvalue $λ = 4$ . Thus we can make a basis of eigenvectors:

β = {(2, - 1, 0), (1, - 1, 1), (1, 4, 16)}

is a Jordan Basis for $A$ . Thus, if we let $S$ be the matrix with those eigenvectors:

S = [\begin{matrix} 1 & 2 & 1 \\ - 1 & - 1 & 4 \\ 1 & 0 & 16 \end{matrix}]

then create $A$ using a change in basis:

A = S [\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 4 \end{matrix}] S^{- 1}

Notice on the diagonal there are the eigenvalues, with the 1's right above for the sub-blocks. But notice that:

A^{n} = S {[\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 4 \end{matrix}]}^{n} S^{- 1}

And using HW 9 - Jordon Form, Complexification#5 we can see that we get:

A^{n} = S {[\begin{matrix} {[\begin{matrix} - 1 & 1 \\ 0 & - 1 \end{matrix}]}^{n} & \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 0 & 0 \end{matrix} & 4^{n} \end{matrix}]}^{n} S^{- 1}

By just raising the blocks to the $n$ -th power, reducing down as far as possible. Because notice for the smaller matrix inside that:

{[\begin{matrix} - 1 & 1 \\ 0 & - 1 \end{matrix}]}^{n} = {([\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}] + [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}])}^{n} = (- I + N)^{n} = \sum_{k = 0}^{n} (- 1)^{n - k} I^{n - k} N^{k} (\binom{n}{k})

But in this case $N^{2} = \vec{0}$ so plug in our $n$ value:

= \sum_{k = 0}^{1} (- 1)^{n - k} (\binom{n}{k}) N^{k} = (- 1)^{n} I + (- 1)^{n - 1} n N

= [\begin{matrix} (- 1)^{n} & (- 1)^{n - 1} n \\ 0 & (- 1)^{n} \end{matrix}]

so then:

A^{n} = S [\begin{matrix} (- 1)^{n} & (- 1)^{n - 1} n & 0 \\ 0 & (- 1)^{n} & 0 \\ 0 & 0 & 4^{n} \end{matrix}] S^{- 1}

So now to bring back to calculate $a_{n}$ we needed to apply $A$ a total of $n$ times:

\begin{aligned} a_{n} & = \underset{select the a_{n} term}{\underset{⏟}{[\begin{array}{c} 1 & 0 & 0 \end{array}]}} A^{n} [\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \end{array}] \\ = [\begin{array}{c} 1 & 0 & 0 \end{array}] A^{n} [\begin{array}{c} - 1 \\ 2 \\ 3 \end{array}] \\ = [\begin{array}{c} 1 & 0 & 0 \end{array}] S [\begin{array}{c} (- 1)^{n} & (- 1)^{n - 1} n & 0 \\ 0 & (- 1)^{n} & 0 \\ 0 & 0 & 4^{n} \end{array}] S^{- 1} [\begin{array}{c} - 1 \\ 2 \\ 3 \end{array}] \\ = \frac{(- 1)^{n} (5 n - 31) + 6 \cdot 4^{n}}{25} \end{aligned}

Some cool things about this is that then the numerator must be divisible by 25 (since the recurrence relation only generates integers, so being divided by 25 implies this).

Application: Solving Systems of Linear DE's

Consider the system ${\vec{x}}^{'} = A \vec{x}$ . It has solution:

e^{A t} = \sum_{n = 0}^{\infty} \frac{(A t)^{n}}{n!} = \sum_{n = 0}^{\infty} A^{n} \frac{t^{n}}{n!}

here the $A^{n}$ part can be more easily calculated by using Jordan-Block form.

Computing Multiplications w/ $(T - λ I)$

Notice if we have:

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

which is $k \times k$ . Multiplying $M (T, β)$ by $M (T - λ I, β)$ is a smaller Jordon block.

Fact

The dimension of the largest jordon block corresponding to $λ$ is the power of $(z - λ)$ in the minimal polynomial.

Example

If we have:

[\begin{matrix} J_{λ, 3} & 0 & 0 \\ 0 & J_{λ, 3} & 0 \\ 0 & 0 & J_{λ, 1} \end{matrix}]

and compare it with:

[\begin{matrix} J_{λ, 3} & 0 & 0 \\ 0 & J_{λ, 2} & 0 \\ 0 & 0 & J_{λ, 2} \end{matrix}]

Notice that these are fundamentally different, representing different operators. However, they have:

the same characteristic polynomial (have 3 + 3 + 1 = 3 + 2 + 2 $λ$ 's on the diagonal, so it has 7 $λ$ ' on the diagonal giving $p_{T} (z) = (z - λ)^{7}$ )
They also have the same minimal polynomail $μ (z) = (z - λ)^{3}$
But they are fundamentally different operators.

Application: Higher Powers of Matrices

Application: Solving Systems of Linear DE's

Computing Multiplications w/ (T−λI)

Example

Computing Multiplications w/ $(T - λ I)$