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I was asked to deduce the Cayley-Hamilton Theorem, that is to show that for all $A \in M_{n}(\mathbb{C})$ we have $\chi_{A}(A)=0$, from the following Corollary:

Let $A$ be an $n \times n$ matrix over $\mathbb{C}$. Then there exists a nonsingular $n \times n$ matrix $P$ over $\mathbb{C}$ s.t. $P^{-1}AP$ is the block matrix

\begin{bmatrix} J(k_{1},\lambda_{1}) & 0 & 0 & \dots & 0 \\ 0 & J(k_{2},\lambda_{2}) & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \dots & J(k_{m},\lambda_{m}) \end{bmatrix}

for some $m \in \mathbb{N}$, $k_{i} \in \mathbb{N} (1 \leq i \leq m)$ s.t. $n=k_{1}+\dots+k_{m}$ and some $\lambda_{i} \in \mathbb{C}$ $(1 \leq i \leq m)$, where $J(k,\alpha)$ will denote the $k \times k$ matrix with $\alpha$'s on the diagonal, $1$'s immediately below and $0$'s elsewhere. Moreover, $\chi_{A}(t)=(t-\lambda_{1})^{k_{1}}\dots(t-\lambda_{m})^{k_{m}}$.

Here is what I did so far:

Let us consider the matrix:

$P^{-1}\chi_{A}(A)P = P^{-1}(A-\lambda_{1}I)^{k_{1}}\dots(A-\lambda_{m}I)^{k_{m}}P=\underbrace{P^{-1}(A-\lambda_{1}I)PP^{-1}\dots PP^{-1}(A-\lambda_{1}I)P}_\text{$k_{1}$ factors $P^{-1}(A-\lambda_{1}I)P$}\dots \underbrace{P^{-1}(A-\lambda_{m}I)PP^{-1}\dots PP^{-1}(A-\lambda_{m}I)P}_\text{$k_{m}$ factors $P^{-1}(A-\lambda_{m}I)P$}=(P^{-1}AP - \lambda_{1}P^{-1}P)^{k_{1}}\dots (P^{-1}AP-\lambda_{m}P^{-1}P)^{k_{m}}=(P^{-1}AP-\lambda_{1}I)^{k_{1}}\dots(P^{-1}AP-\lambda_{m}I)^{k_{m}}=? $

Now, I'm kind of stuck here. I suspect that this should be equal to $0$, which if true would give me my result, but I'm not really sure how to show that, assuming that it is indeed the case. Any ideas?

amator2357
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  • I don't really understand your comment. The image under what map? What are the $v$'s? And how would that help me prove my claim? – amator2357 Feb 08 '19 at 13:32

1 Answers1

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You are trying to show that:

$(P^{-1}AP-\lambda_{1}I)^{k_{1}}\dots(P^{-1}AP-\lambda_{m}I)^{k_{m}}=0$

let's take one subspace: $(P^{-1}AP-\lambda_{1}I)^{k_{1}}$ then it's equal to \begin{bmatrix} J(k_{1},0) & 0 & 0 & \dots & 0 \\ 0 & J(k_{2},\lambda_{2}-\lambda_{1}) & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \dots & J(k_{m},\lambda_{m}-\lambda_{1}) \end{bmatrix}

now we see that each group is distinct in the matrix so if we take a vector belonging to the first bloc we have a reduced computation: $$ (P^{-1}AP-\lambda_{1}I)^{k_{1}}X = \begin{bmatrix} J(k_{1},0) & 0 & 0 & \dots & 0 \\ 0 & J(k_{2},\lambda_{2}-\lambda_{1}) & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \dots & J(k_{m},\lambda_{m}-\lambda_{1}) \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{bmatrix} $$

now we look only at $k_1$ first coordinates and we will show that this going to 0.

$$ (P^{-1}AP-\lambda_{1}I)^{k_{1}}X = $$ $$ \begin{bmatrix} J(k_{1},0)^{k_{1}} [x_1,..,x_{k_1}] & 0 & 0 & \dots & 0 \\ 0 & J(k_{2},\lambda_{2}-\lambda_{1})^{k_{1}}[x_{k_1+1},..,x_{k_1+k_2}] & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \dots & J(k_{m},\lambda_{m}-\lambda_{1})^{k_{1}}[x,..x_n] \end{bmatrix} $$

if $J(k_{1},0) [x_1,..,x_{k_1}]$ = 0 then you can see that it stays true for all i and the result comes from it because each $J$ is putting a part of the vector to 0.

You just now have to show that $J(k_{i},0)$ is nilpotent of order $k_i$ and you are good.

Frayal
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