Two Diagonalizable Matrices Have Common Diagonalisation when $AB=BA$

Question

The following is an exercise:

Thm.: $A_{n \times n}$ is diagonalizable iff its minimal polynomial $m_A(t)$ splits into distinct linear factors.

Use this theorem to prove that if $A, B \in \mathbb{F}^{n \times n}$ are diagonalizable matrices such that $AB=BA$ then there exists a $P$ such that $P^{-1}AP,\, P^{-1}BP$ are diagonal matrices. In other words, $A$ and $B$ have $n$ common linearly independent eigenvectors.

Could anyone give me a hint on how to start this proof?

Answer 1

I'll answer in the language of linear maps which can be translated easily to the required result for matrices. If $T \colon V \rightarrow V$ is diagonalizable and $W \subseteq V$ is $T$-invariant then $T|_W$ is also diagonalizable. The reason is that the minimal polynomial of $T|_W$ must divide the minimal polynomial of $T$ and so if the minimal polynomial of $T$ splits into distinct linear factors, so does the minimal polynomial of $T|_W$ which implies that $T|_W$ is diagonalizable.

Now, if $T,S \colon V \rightarrow V$ are diagonalizable and commute ($TS = ST$), let $\lambda_1, \dots, \lambda_k$ denote the distinct eigenvalues of $T$ and by $V_i = \ker(T - \lambda_i I)$ the corresponding eigenspaces. If $v \in V_i$ then $Tv = \lambda_i v$ and $T(Sv) = S(Tv) = S(\lambda_i v) = \lambda_i Sv$ so $Sv \in V_i$. Thus, each $V_i$ is $S$-invariant and since $S$ is diagonalizable, so is $S|_{V_i}$ for each $1 \leq i \leq k$. If we choose a basis of eigenvectors in $V_i$ for $S|_{V_i}$ for each $1 \leq i \leq k$ we will obtain a basis of eigenvectors of both $S$ and $T$ which will imply that $S$ and $T$ are simultaneously diagonalizable.

Answer 2

(I'm sure that anyone can translate this to english. My english is very poor).

-> Supose that $h$,endomorphism asociated to $A$ matrix, is diagonalizable. Then, it exists, $K$-basis $\{a_1, \ldots, a_n\}$ of $V$, composes by eigenvectors.

Then, for all $i\in\{1,\ldots, n\}$ exists $\lambda_i \in K$ : $$h(a_i)=\lambda_i a_i$$

I.e. $$x a_i = \lambda_i a_i$$ I.e. $$(x-\lambda_i) a_i = 0$$

This means that $$pol min a_i = x-a_i \text{ for all } i\in\{1,\ldots, n\}$$

Because of $V=<a_1,\ldots, a_n>$, it is true tha $$pol min h = pol min V = mcm\{x-\lambda_i : i\in\{1,\ldots, n\} \}$$

So, $$pol min h = (x - \lambda_1) \cdot \ldots \cdot (x- \lambda_t)$$ where $\{\lambda_1, \ldots, \lambda_t\}$ are the diferent eigenvectors of $h$.

Then, all invariant factor of $h$ is the product of diferent polinomials of grade 1. So, all elemental divisor of $h$ has grade 1.

<- Trivial because the canonic form of Jordan is diagonal.

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-> Supongamos que $h$, endomorfismo asociado a la matriz $A$ es diagonalizable. Entonces, existe una $K$-base, $\{a_1, \ldots, a_n\}$ de $V$ formada por vectores propios.

Entonces, para todo $i\in\{1,\ldots, n\}$ existe $\lambda_i \in K$ de forma que $h(a_i)=\lambda_i a_i$. Es decir, $x a_i = \lambda_i a_i$. En otras palabras, $(x-\lambda_i) a_i = 0$. Esto implica que $pol min a_i = x-a_i$ para todo $i\in\{1,\ldots, n\}$.

Como $V=<a_1,\ldots, a_n>$ tenemos que $pol min h = pol min V = mcm\{x-\lambda_i : i\in\{1,\ldots, n\} \}$. Por tanto, $$pol min h = (x - \lambda_1) \cdot \ldots \cdot (x- \lambda_t)$$ donde $\{\lambda_1, \ldots, \lambda_t\}$ son los valores propios diferentes de $h$. Entonces, todo factor invariante de $h$ es producto de polinomios de grado 1 diferente. Concluimos que todo divisor elemental de $h$ es de grado 1.

<- Trivial puesto que la forma canónica de Jordan, en este caso, es diagonal.