# The minimal polynomial written as a product

For $A \in \mathbb{C}^{n,n}$ and $\{ \lambda_1, \dots , \lambda_r\}$ are the eigenvalues of $A$.

My lecture notes say that the minimal polynomial of $A$ is $$\prod_{i=1}^r(x-\lambda_i)^{a_i}$$ where $a_i$ is the largest among the degrees of the Jordan blocks of $A$ of eigenvalue $\lambda_i$.

I can't get my head around how this is the case? Can someone try and explain this to me?

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What exactly are you aiming at? What would you expect it to be? –  user20266 Apr 24 '12 at 14:06
Are you having trouble going from "sizes of the Jordan blocks" to "this is what the minimal polynomial is", from "this is the minimal polynomial" to "then the largest size of the Jordan blocks is this", or both directions? –  Arturo Magidin Apr 24 '12 at 15:41
Both directions really –  user26069 Apr 24 '12 at 16:45

## 1 Answer

The rough outline of the proof is:

1. If $A$ and $T$ are square matrices, and $T$ is invertible, then the minimal polynomial for $A$ is the same as the minimal polynomial for $TAT^{-1}$.

2. If $B$ is a square matrix composed of a series of square matrices, $B_1,...,B_k$ along the diagonal and zero elsewhere, then the minimal polynomial for $B$ is the least common multiple of the minimal polynomials for the $B_i$.

3. If $J_m(\lambda)$ is an $m\times m$ Jordan block - $\lambda$ on the diagonal and $1$ just above the diagonal - then the minimal polynomial for $J_m(\lambda)$ is $p(x)=(x-\lambda)^m$.

So, for any $A$, you can find a $T$ so that $B=TAT^{-1}$ is in Jordan Normal Form. We know that the minimal polynomial for $B$ is the least common multiple of the Jordan blocks of $B$. And we know the minimal polynomials for the Jordan blocks, which yields the result.

(1) is easy to prove.

(2) is easy if you realize that if $B$ is of this form, and $p$ is any polynomial, then $p(B)$ is composed of $p(B_1),...,p(B_k)$ along the dialogal and zero elsewhere.

(3) Takes a little bit of arithmetic. First prove it for $\lambda=0$.

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