If $AB=BA$, prove that $ A=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} $

linear algebramatrices

Let $A$ a $2\times2$ matrix, if $AB=BA$ for every $B$ of the size $2\times2$, Prove that:

$$
A=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}
$$

$a \in \mathbb{R}$

My attempt:

Let $$
A=\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix}
$$

$$B=\begin{bmatrix} a_2 & b_2 \\ c_2 & d_2 \end{bmatrix}$$

And since $AB=BA$, then

$a_1 a_2 + b_1 c_2 = a_1a_2+b_2 c_1$

So $b_2 c_1=b_1 c_2$

And

$a_1 b_2+b_1 d_2=a_2 b_1+b_2 d_1$

$c_1 a_2+d_1 c_2=c_2 a_1+d_2 c_1$

But what can I do now ?
Thanks 🙂

Best Answer

Let $A=\pmatrix{a&b\\c&d}$. For $B=\pmatrix{1&0\\0&0}$ we have $AB=\pmatrix{a&0\\c&0}$ and $BA=\pmatrix{a&b\\0&0}$. So $AB=BA$ implies $b=c=0$, that is $A=\pmatrix{a&0\\0&d}$. Now try, say $B=\pmatrix{0&1\\0&0}$.

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