Show that every eigenvector of A is an eigenvector of B

Question

Let $A, B\in\mathcal M_n(\mathbb{R})$ such that $AB=BA$ with $n$ distinct eigenvalues.

1) Show that if $v\in\mathbb{R}^n$ and $\lambda\in\mathbb{R}$, $Av=\lambda v\implies ABv=\lambda Bv$

2) Show that every eigenvector of $A$ is an eigenvector of $B$

3) Show that $B$ is diagonalisable.

4) Show that $AB$ is diagonalisable.

My attempt:

1) We have that: $Av=\lambda v$ Let's multiply to the left with $B$.

So now we have: $BAv=B\lambda v$, but $BA=AB\implies ABv=\lambda Bv$ because $\lambda$ is in $\mathbb{R}.$

since $A$ has n distinct eigenvalues $\implies$ $A$ is diagonalisable.

How do I show that $B$ is too diagonalisable? and that every eigenvector of $A$ is an eigenvector of $B$?

Answer 1

Let $Av=\lambda v$. Then

$$Bv=\frac1{\lambda}B\lambda v=\frac1{\lambda}BAv=\frac1{\lambda}ABv$$ Hence, if we set $x:=Bv$, we can rewrite this as $Ax=\lambda x$. Now, using that $A$ has $n$ distinct eigenvectors, and hence, that each eigenspace is $1$-dimensional, we infer that $x$ must be linearly independent with $v$, i.e., there exists $k\in\mathbb R$ such that $x=kv$ or equivalently, $Bv=kv$.

Iterating this, you obtain that $B$ has $n$ distinct eigenvalues as well. So, $B$ is diagonalizable.

Answer 2

Let $v\in E_\lambda(A)$ be an eigenvector of $A$, then $ABv=\lambda Bv$ and $Bv$ is a $\lambda$-eingenvector too.

Since $A$ has $n$ distinct eigenvalues, $dim(E_\lambda(A))=1$ and $\exists \mu$ so that $Bv=\mu v$.

Answer 3

If a Matrix $A$ is diagonalizable, there will be a Diagonal Matrix $D_A$ that has a simular condition:

$D_A = S^{-1}AS$

You know that $A$ is diagonalizable if there is a transformation matrix $S\in\mathbb{C}^{n\times n}$

In order to diagonalize the matrix you need to find $D_A$

If you know that $B$ has the same transformation matrix $S$ you can diagonalize both of them. If this is possible you have:

$S^{-1}AS=D_1$

and

$S^{-1}BS=D_2.$

Then you know that D1 and D2 are diagonal matrices.

Thus:

$D_1\cdot D_2 = D_2\cdot D_1 \Rightarrow B\cdot A= SD_2S^{-1}\cdot SD_1S^{-1}= SD_1D_2S^{-1}= A\cdot B$

You need to know how to determine the eigenvalues for this.