Eigenvalues of $A$ Let  $A$ be a $3*3$  matrix with real entries. If $A$ commutes with all $3*3$ matrices with real entries, then how many number of distinct real eigenvalues exist for $A$?
please give some hint.
 thank you for your time.
 A: Hint: denote by $E_{ij}$ the matrix with all zeroes, except for the $ij$-th entry which is $1$. Write out what does it mean for $A$ to commute with $E_{ij}$ - i.e. just multiply by it from the right, from the left and compare.

Completely different approach: Let $\mu,\lambda$ be two eigenvalues of $A$ corresponding to (non-zero) eigenvectors $v,w$ respectively. Find a matrix $B$ such that $Bv=w$, $Bw=v$. Now compute $ABv$ and $BAv$. They should be equal, since $AB=BA$
A: If $A$ commute with $B$ then $$[A,B] = AB - BA = 0$$
$$AB=BA$$ 
$$ABB^{-1}=BAB^{-1}$$ 
$$A=BAB^{-1}$$
If the $B$ has inverse matrix $B^{-1}$, then sets of eigenvalues not change.
Matrix $3\times{3}$ has 3 eigenvalues.
A: The value $3$ in the question is of no importance (as long as it is not $0$).
If matrices $A,B$ commute, then $A$ must stabilise $\ker B$: if $x\in \ker B$ and $y=A\cdot x$ then $B(y)=B\cdot A(x)=A(B(x))=A(0)=0$ and so $y\in\ker B$.
Now choosing $B$ appropriately $\ker B$ can be any subspace you want, so $A$ stabilises all subspaces. In particular it stabilises $\langle v\rangle$ for any nonzero vector $v$, so $A(v)\in\langle v\rangle$ and $v$ is an eigenvector; in particular (since we are in a real vector space) $A$ has some real eigenvalue$~\lambda$. But then since $A-\lambda I$ commutes with every $B$, the nonzero subspace $\ker(A-\lambda I)$ is a stable by all matrices, which can only mean it is the entire space. So $A=\lambda I$  does not have any other eigenvalues.
A: Hint:
Show that $A=\lambda I$ for some $\lambda\in\mathbb R$.
