# proof that if $AB=BA$ matrix $A$ must be $\lambda E$ [duplicate]

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?

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## marked as duplicate by Marc van Leeuwen, Dennis Gulko, Daniel Fischer♦, Davide Giraudo, Nick PetersonOct 30 '13 at 13:36

Since $AB=BA$ holds for all $B$, try putting in some simple matrices for $B$ and see what you can conclude about $A$. Simple means a matrix that contains many 0's (but not all 0's because then you just get 0=0 which doesn't tell you anything). – Ted Feb 4 '12 at 19:09
Take $B$ to be each of the basis matrices $E_{ij}$ that is all zeros except that it is 1 at position $ij$.