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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 $B$?

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  • $\begingroup$ See this and that or any question linked to those. $\endgroup$ – Arnaud D. Nov 6 '18 at 15:38
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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.

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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 $Bv=\mu v$.

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  • $\begingroup$ Why $Bv$ is a $\lambda-$eigenvector too? $\endgroup$ – C. Cristi Nov 6 '18 at 15:29
  • $\begingroup$ @C.Cristi $A(Bv) = \lambda (Bv)$. $\endgroup$ – xbh Nov 6 '18 at 16:24
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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.

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  • $\begingroup$ A common conclusion from the question in the OP is that $A, B$ could be diagonalized by the same matrices. So you kind of use the conclusion to prove the assumption. The answer you gave does not give much information about solving the question, personally speaking. $\endgroup$ – xbh Nov 6 '18 at 16:28

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