A be a $3\times 3$ matrix over $\mathbb {R}$ such that $AB =BA$ for all matrices $B$. what can we say about such matrix $A$ Let $A$ be a $3\times 3$ matrix over $\mathbb {R}$ such that $AB =BA$ for all matrices $B$ over $\mathbb {R}$ then what can we say about such matrix $A$. 
or such matrix $A$ must be orthogonal matrix? Can we say anything about its eigen values?
 I tried by taking random examples also tried to construct such $3 \times 3$ matrices. But i am not able to get any proper conclusion and proof. 
 A: Here is one way to prove this, a bit less elementary than what was suggested in the comments (but it avoids any computations). It is a fundamental fact (and easy to prove) that if $A,B$ are commuting matrices, then every eigenspace for $B$ must be $A$-stable (i.e., if $v$ is in the eigenspace of$~B$ for eigenvalue$~\lambda$, then so is $A\cdot v$). Also, a $1$-dimensional subspace $\langle v\rangle$ is $A$-stable precisely if $v$ is an eigenvector for $A$ (immediate from the definition).
On one hand if $A$ is a scalar multiple of identity, then it clearly commutes with every $B$ (by definition any linear operator commutes with scalar multiplications). On the other hand if $A$ is not a scalar multiple of identity, then there exists some vector $v$ that is not an eigenvector of $A$ (if every nonzero vector were eigenvector, one easily shows they all have the same eigenvalue $\lambda$, and $A$ would be $\lambda I$). Then choose $B$ such that $\langle v\rangle$ is an eigenspace for $B$ (easy). By the above, if $B$ would commute with $A$, then $\langle v\rangle$ would be $A$-stable, and therefore $v$ an eigenvector for $A$. But it isn't, so $A$ and $B$ do not commute.
Note that the facts that the base field is $\Bbb R$, that the dimension is$~3$ or even that it is finite are irrelevant to this question.
A: Here is a simple exercise which I find useful few times.
Exercise Let $D$ be a diagonal matrix, with pairwise distinct diagonal entries. Then $D$ commutes only with diagonal matrices.
Proof: Let $CD=DC$. Equaling the $ij$ entries in these matrices we get:
$$c_{ij}d_{jj}=d_{ii}c_{ij}$$
Thus, if $i \neq j$we have $d_{ii} \neq d_{jj}$ and thus $c_{ij}=0$..
Now back to your problem. Pick $B$ a diagonal matrix with pairwise distinct diagonal entries. Then $AB=BA$ implies that $A$ is diagonal.
Next pick $B$ the matrix with all ones, that is $b_{ij}=1 \forall i,j$. $AB=BA$  and $A$ diagonal implies that all entries of $A$ are equal.
A: Here's another solution. Since $A$ is $3\times3$ and $3$ is odd, $A$ has a real eigenvalue $\lambda$ (if you're comfortable with complex eigenvalues, we don't even need the dimension to be odd.) We claim that $A=\lambda I$. To prove this, let $w = (w_1,w_2,w_3)$ be an arbitrary vector in $\mathbb R^3$; it suffices to show that $wA =\lambda w$.
Let $v$ be an eigenvector corresponding to $\lambda$. Some coordinate of $v$ must be nonzero, and $v$ can be scaled at will; without loss of generality, say the first coordinate of $v$ equals $1$. Now set $B$ to be the matrix whose columns are $w_1v$, $w_2v$, and $w_3v$. We easily compute that $AB$ is the matrix whose columns are $\lambda w_1v$, $\lambda w_2v$, and $\lambda w_3v$; in particular, the top row of $AB$ equals $(\lambda w_1,\lambda w_2,\lambda w_3) = \lambda w$. This is also the top row of $BA$, which equals the top row of $B$ times $A$, which equals $w$ times $A$. This establishes that $wA =\lambda w$ as desired.
