Lecture 12

[[lecture-data]]

Announcement

Exam in about 2 weeks - choose the “least bad” date.

2024-09-23

Readings

• a

3. Chapter 3

Jordan’s theorem

Recall the statement of Jordan’s Theorem from last Lecture Lecture 11. Today we do the proof: uniqueness (+existence maybe later or just reading)

Theorem

For each $A\in M_n$ there exists a Jordan Matrix $J$ such that $A$ is similar to $J$: $A=SJ{S}^{-1}$ for $S\in M_n$. Further, $J$ is unique up to reordering its Jordan blocks.

$A\in M_3$ with all eigenvalues $\pi$ has 3 possible similarity classes, one for each size of largest Jordan blocks.

(see Jordan’s Theorem)

Note

When we say Jordan block we mean having eigenvalues along the diagonal and all $1$s on the superdiagonal.

Proof (uniqueness)

Consider a special case, $A\in M_n$ which has only $0$s as eigenvalues. $A=SJ{S}^{-1}$ where $J={\oplus }_{i=1}^s{J}_{n_2}(0)$. What is $\text{rank}{A}^0$ ? $\text{rank}{A}^0=n_1+n_2+\cdots +n_s$ Note that $\text{rank}{A}^k=\text{rank}{J}^k$ and recall that $\text{rank}{J}^k={\sum }_{i=1}^s(n_i-k)$ (this is for when the all eigenvalues are $0$ !)

Dual Sequence

Consider a sequence of non-increasing natural numbers. This can be considered a partition of their sum.

$(5,3,3,2)$

We can draw a Ferrer’s Diagram of this sequence. And we can take the “transpose” or look at the number of blocks on the left side. And this gives another partition of the sum.

The dual sequence is “counting the blocks” on the left side.

$(5,3,3,2{)}^{\ast }=(4,4,3,1,1)$

So order the $n_i$ of the Jordan Matrix for $A$ in a non-increasing sequence. Then

Note

$\text{rank}{A}^{\ell -1}-\text{rank}{A}^{\ell }=\ell$th entry of the dual sequence to block sequence.

If $A\in M_n$ has only eigenvalues zero, define $s_{\ell }:=\text{rank}{A}^{\ell -1}-\text{rank}{A}^{\ell }$. Then $(t_1,t_2,\dots )=(s_1,s_2,\dots {)}^{\ast }$ is the block structure sequence, ie $A\sim {\oplus }_{i=1}{J}_{t_i}(0)$ ie, we can deduce the block structure from $A$ itself.

More generally, suppose $A\in M_n$ where $A\sim S[{\oplus }_{i=1}{J}_{n_i}({\lambda }_i)]S^{-1}$. How do I convert to the case before? Consider

A - \gamma I &= S[\oplus_{i}J_{n_{i}}(\lambda_{i}) - \gamma I]S^{-1},\;\;\;\; \gamma \in \mathbb{C} \\ &= S[\oplus_{i}J_{n_{i}}(\lambda_{i} - \gamma) I]S^{-1} \end{aligned}$$ What if we take $\gamma = \lambda_{1}$ ? Then $$\begin{bmatrix} J_{n_{1}}(0) & & & & & \\ & J_{n_{2}}(0) & & & & \\ & & \ddots & & & & \\ & & & J_{n_{q}}(\neq 0) & & \\ & & & & J_{n_{q+1}}(\neq 0) & \\ & & & & & \ddots \end{bmatrix}$$ So we have that $\text{rank}(A - \gamma I)^\ell = \text{rank}(J-\gamma I)^\ell$ $$= \text{rank}\begin{bmatrix} J_{n_{1}}(0)^\ell & & & & & \\ & J_{n_{2}}(0)^\ell & & & & \\ & & \ddots & & & & \\ & & & J_{n_{q}}(\neq 0)^\ell & & \\ & & & & J_{n_{q+1}}(\neq 0)^\ell & \\ & & & & & \ddots \end{bmatrix}$$ And the blocks $$\begin{bmatrix} J_{n_{q}}(\neq 0)^\ell & & \\ & J_{n_{q+1}}(\neq 0)^\ell & \\ & & \ddots \end{bmatrix}$$ are invertible! Since we know the eigenvalues are not zero. As we increase $\ell$, the rank of top of the block is going to decrease as we saw above, but the rank of the entire matrix will always include the rank of the invertible block. - we can observe the difference in the rank of the matrix for the successive powers of $A$, and that will tell us the sizes of the blocks for the first eigenvalue - We can then repeat this process with each eigenvalue to recover the block structure associated with each eigenvalue of $A$, and thus the form of $A$! >[!theorem] Claim > >If $A \in M_{n}$ and we know $\sigma(A)$, you can reconstruct the [[Concept Wiki/Jordan's Theorem\|Jordan canonical form]] from >$\forall \lambda \in \sigma(A), \;\;\text{rank}(A - \lambda I)^\ell \;\;\;\forall \ell=0, 1, 2, \dots$ For $A \in M_{n}$ similar to $J=\oplus_{i=1}^k J_{n_{i}}(\lambda_{i})$ such that $\lambda_{1}=\lambda_{2}=\lambda_{3}=\dots=\lambda_{s} \neq \lambda_{i} \;\;\;\forall i >s$. The [[Concept Wiki/spectrum\|geometric multiplicity]] of $\lambda_{1}$, call it $\mu_{g}=1+1+\dots+1 = s$ , and the [[Concept Wiki/spectrum\|algebraic multiplicity]] of $\lambda_{1}$, call it $\mu_{a} = n_{1}+n_{2}+n_{3}+\dots+n_{s}$. In particular, for any eigenvalue, the geometric multiplicity has the property that $1 \leq \mu_{g} \leq \mu_{a}$. >[!example] > >$A = SJS^{-1}$. Suppose $J$ has a 3 [[Concept Wiki/Jordan block\|block]] with $\pi$, a 2 block with $\pi$ and a 2 block with $e$. Then the dimension of the [[Concept Wiki/eigenvalue\|eigenspace]] with [[Concept Wiki/eigenvalue]] $\pi$ is the [[Concept Wiki/nullspace\|nullity]] of $\pi I -A$. >$$\begin{aligned} >\text{nullity}(\pi I - A) &= \text{nullity}(\pi I- SJS^{-1}) \\ >&= \text{nullity} S (\pi I -J)S^{-1} \\ >&= \text{nullity}(\pi I - J) \\ >&= \text{nullity} \begin{bmatrix} > 0 & -1 & & & & & > \\ & 0 & -1 & & & & > \\ & & 0 & & & & > \\ & & & 0 & -1 & & > \\ & & & & 0 & & > \\ & & & & & \pi-e & > \\ & & & & & & \pi-e >\end{bmatrix} >\end{aligned}$$ > >>[!exercise] >>So what is the geometric multiplicity of $\pi$? ie, what is the dimension of the eigenspace? > > >[!theorem] > >Suppose $A \in M_{n}$. Then $A \sim A^T$. >>[!proof]+ >>We know that $p_{A}(t) = p_{A^T}(t)$. For any $\lambda \in \sigma(A) [=\sigma(A^T)]$ then >>$$\text{rank}(A - \lambda I)^\ell = [\text{rank}(A - \lambda I)^\ell]^T = \text{rank}[A^T - \lambda I]^\ell$$ >>By the theorem above, this means that we get the same [[Concept Wiki/Jordan's Theorem\|Jordan canonical form]] for each of them, meaning they are [[Concept Wiki/similar matrices\|similar]] to the same matrix. And since similarity is an equivalence relation, they are similar to each other as well! ![[Attachments/pasted/FullSizeRender.jpg]]