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Let $A$ a $2x2$ matrix, if $AB=BA$ for every $B$ of the size $2x2$, 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 :)

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  • $\begingroup$ This works for $n\times n$ matrices in general. In fact, if $D$ is a division ring, then the center of $M_n(D)$ consists of the scalar matrices $d\cdot I_n$, where $I_n$ is the identity matrix and $d$ is an element of the center of $D$. $\endgroup$ – A. Pongrácz Jul 19 '18 at 12:55
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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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  • 2
    $\begingroup$ Thank you so much,when I take $B=\pmatrix{0&1\\0&0 I got $a=d$, then we proved it :) $\endgroup$ – Dima Jul 19 '18 at 11:41
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Note: $$\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}= \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \Rightarrow \\ \begin{cases}\require{cancel}\cancel{a_{11}b_{11}}+a_{12}b_{21}=\cancel{b_{11}a_{11}}+b_{12}a_{21}\\ a_{11}b_{12}+a_{12}b_{22}=b_{11}a_{12}+b_{12}a_{22}\\ a_{21}b_{11}+a_{22}b_{21}=b_{21}a_{11}+b_{22}a_{21}\\ a_{21}b_{12}+\cancel{a_{22}b_{22}}=b_{21}a_{12}+\cancel{b_{22}a_{22}}\end{cases}$$ From $(1)$, since $b_{12}$ and $b_{21}$ can be any number, in particular, $b_{12}=0$ and $b_{21}\ne 0$, we get: $a_{12}=0$.

Similarly, for $b_{12}\ne 0$ and $b_{21}=0$, we get $a_{21}=0$.

From $(2)$, since $a_{12}=0$ and $b_{12}$ is an arbitrary number, we get $a_{11}b_{12}=b_{12}a_{22} \Rightarrow a_{11}=a_{22}$.

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