Prove that if $B$ is a $2 × 2$ matrix commuting with $A$, then $B$ is a diagonal matrix

Question

I have the following statement I need to prove:

Suppose $A = \left[ \begin{array}{cc} a&0\\ 0&b\\ \end{array} \right]$ is a diagonal matrix, where $a ≠ b$. Prove that if $B$ is a $2 × 2$ matrix commuting with $A$, then $B$ is a diagonal matrix.

My proof:

In particular, let $B = \left[ \begin{array}{cc} c&0\\ 0&d\\ \end{array} \right]$.

Suppose $B$ is not a diagonal matrix, then the commutative property of diagonal matrices, that is $AB = BA$, should not hold.

We see that $AB = \left[ \begin{array}{cc} ac&0\\ 0&bd\\ \end{array} \right]$ and $BA = \left[ \begin{array}{cc} ca&0\\ 0&bd\\ \end{array} \right]$ which is basically $ \left[ \begin{array}{cc} ac&0\\ 0&bd\\ \end{array} \right]$.

Since $AB = BA$ this is a contradiction. Thus $B$ is a diagonal matrix.

END OF PROOF

I feel like I'm completely off here. Any guidance would be appreciated.

EDIT: What I got with all the help.

Suppose $A = \left[ \begin{array}{cc} a&0\\ 0&b\\ \end{array} \right]$ is a diagonal matrix where $a ≠ b$.

Let B be an arbitrary matrix $\left[ \begin{array}{cc} w&x\\ y&z\\ \end{array} \right]$.

By definition, for matrices to be commuting AB = BA must hold.

We see that AB $\left[ \begin{array}{cc} aw&ax\\ by&bz\\ \end{array} \right]$ and BA = $\left[ \begin{array}{cc} aw&bx\\ ay&bz\\ \end{array} \right]$.

For these two matrices to be equal x = y = 0 must be true.

It then follows that B = $\left[ \begin{array}{cc} w&0\\ 0&z\\ \end{array} \right]$ which is indeed a diagonal matrix.

Is this all I need?

Answer 1

The matrix $$\pmatrix{c&0\cr0&d\cr}$$ is diagonal. So it makes no sense to say "let $B$ be this matrix, suppose $B$ is not diagonal". That's just like saying "let $x=100$, suppose $x$ is not positive".

To solve the problem, take $$B=\pmatrix{w&x\cr y&z\cr}\ ,$$ multiply out $AB=BA$ in terms of components, and prove $x=0$ and $y=0$. Don't forget the fact that $a\ne b$.

Good luck!

Answer 2

You write $B$ as a diagonal matrix to start with, so this is no good. What you want to do is start with an arbitrary $B$ and show that it must be diagonal if it commutes with $A$, where $A$ is diagonal with distinct diagonal components. So write $$B=\begin{bmatrix}a&b\\c&d\end{bmatrix}\qquad\text{and}\qquad A=\begin{bmatrix}x&0\\0&y\end{bmatrix}~,~x\neq y$$ then show that $AB=BA$ implies $c=d=0$ to show that $B$ is diagonal.

Answer 3

Take a more general case for B matrix where all the elements are different and then prove that elements on non primary diagonal are equal to zero using the conditions given in question .