# Conditions for symmetry for pair of commuting matrices

Let A = $\left(\begin{array}{ccc} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} \right)$

Show that every real matrix $B$ such that $AB = BA$ has the form $B = aI + bA + cA^2$

My attempt: If we assume that $AB = BA$ are simultaneously diagonalizable by $P$, then since $A = PDP^{-1}$we have the RHS expression to be $aPIP^{-1} + bPDP^{-1} + cPD^2P^{-1}$. Computing $D$, we observe that $I$ and $D$ and $D^2$ are linearly independent, and thus they span all 3X3 diagonal matrices, diagonalized B included.

But $A$ and $B$ are simultaneously diagonalizable iff the commute and they are both diagonalizable. Now $A$ is real symmetric so it is diagonalizable, but what can we say about $B$?

It looks like $B$ wants to be symmetric. But A symmetric and $AB=BA$ does not imply B is symmetric, take $A = I$ for instance. So I have two questions:

1) Under what conditions for A will the statement ($A$ symmetric and $AB= BA) \Rightarrow B$ symmetric hold?

2) If my approach does not work, does anyone have any other way of proceeding?

This particular $A$ has distinct eigenvalues (it's not hard to factor the characteristic polynomial). Every matrix $B$ that commutes with $A$ must preserve the eigenspaces, thus the eigenvectors of $A$ must be eigenvectors of $B$, and in particular $B$ is diagonalizable.

If a symmetric matrix does not have distinct eigenvalues, there will be matrices that commute with it but are not diagonalizable. For example, if $v$ and $w$ are orthogonal eigenvectors of $A$ for the same eigenvalue, $v w^T$ is such a matrix (note that its square is $0$).

• That was most helpful. Thanks. Nov 21 '15 at 9:17

We can prove the statement by brute force.

Search a matrix $B$ the commute with $A$:

$$\begin{bmatrix} 2&-1&0\\ -1&2&-1\\ 0&-1&2 \end{bmatrix} \begin{bmatrix} x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3 \end{bmatrix}= \begin{bmatrix} x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3 \end{bmatrix} \begin{bmatrix} 2&-1&0\\ -1&2&-1\\ 0&-1&2 \end{bmatrix}$$ wit a bit of work, equating the corresponding entries of the two products, we find that $B$ has the form: $$B=\begin{bmatrix} x&y&z\\ y&x+z&y\\ z&y&x \end{bmatrix}$$

Now: $$aI+bA+cA^2= a\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}+b \begin{bmatrix} 2&-1&0\\ -1&2&-1\\ 0&-1&2 \end{bmatrix}+c \begin{bmatrix} 5&-4&1\\ -4&6&-4\\ 1&-4&5 \end{bmatrix}= \begin{bmatrix} 2a+2b+5c&-b-4c&c\\ -b-4c&a+2b+6c&-b-4c\\ c&-b-4c&a+2b+5c \end{bmatrix}$$ tha has the form of $B$ for $x=2a+2b+5c$, $y=-b-4c$ and $z=c$.