<s>[INST] proof that if $AB=BA$ matrix $A$ must be $\lambda E$ Let $A \in Mat(2\times 2, \mathbb{Q})$ be a matrix with $AB = BA$ for all matrices $B \in Mat(2\times 2, \mathbb{Q})$.
Show that there exists a $\lambda \in \mathbb{Q}$ so that $A = \lambda E_2$.

Let $E_{ij}$ be the matrix with all entries $0$ except $e_{ij} = 1$.
$$
\begin{align}
AE_{11} &= E_{11}A \\
\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array} \right)
\left( \begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array} \right)
&=
\left( \begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array} \right)
\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array} \right)
\\
\left( \begin{array}{cc}
a_{11} & 0 \\
a_{21} & 0 \\
\end{array} \right)
&=
\left( \begin{array}{cc}
a_{11} & a_{12} \\
0 & 0 \\
\end{array} \right) \\
\end{align}
$$
$\implies a_{12} = a_{21} = 0$
And then the same for the other three matrices $E_{12}, E_{21}, E_{22}$ …
I guess it's not the most efficient way of argueing or writing it down … ? Where's the trick to simplify this thingy?
 [/INST] Take $B$ to be each of the basis matrices $E_{ij}$ that is all zeros except that it is 1 at position $ij$.
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