# 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?

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

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• 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
• you may interested to check Schur's lemma. – Kerry Feb 4 '12 at 20:00

## 2 Answers

Take $B$ to be each of the basis matrices $E_{ij}$ that is all zeros except that it is 1 at position $ij$.

You wish to show that the center of the ring of matrices is the set of scalar matrices. Observe that the matrices commuting with diagonal matrices must be diagonal matrices. Hence the center must be contained in the set of diagonal matrices. Now a diagonal matrix that commutes with an arbitrary matrix must be a scalar matrix.