# Cayley-Hamilton Theorem proof

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?

• 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? Feb 8, 2019 at 13:32

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.

• Thank you @Alexis! I can't quite see why this equality: $(P^{-1}AP-\lambda_{1}k)^{k_{1}}X=J(k_{1},0)^{k_{1}}X_{J_{1}}$ holds? Could you explain, please? Feb 8, 2019 at 14:08
• i ll update the answer Feb 8, 2019 at 14:13
• Thank you! Much appreciated. Feb 8, 2019 at 14:16
• i'm not sure if this is clear enough Feb 8, 2019 at 14:32
• Ok, I think I get it now. So basically, we show that $J(k_{i},0)$ is nilpotent of order $k_{i}$ and what follows is that $J(k_{i},0)$ will put $J(k_{i+1},\lambda_{1} - \lambda_{2})$ to $0$ and so on and so we'll end up zeroing all the entries of $(P^{-1}AP-\lambda_{1}I)^{k_{1}}\dots (P^{-1}AP-\lambda_{m}I)^{k_{m}}$? Feb 8, 2019 at 15:12