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 $B$?
Best Answer
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.