Commutative matrix proof I've got the following question:
$A \in M_{nn}(\mathbb{K})$ is a matrix and $AB=BA \forall B \in M_{nn}$. Proof that $ A=aI_n \forall a \in \mathbb{K}$.
and one given solution starts with:
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{E_{ii}}A = \left( {\begin{array}{*{20}{c}}
0& \cdots &0& \cdots &0\\
 \vdots &{}& \vdots &{}& \vdots \\
{{a_{i1}}}& \cdots &{{a_{ii}}}&{}&{{a_{in}}}\\
 \vdots &{}& \vdots &{}& \vdots \\
0& \cdots &0& \cdots &0
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
0& \cdots &{{a_{1i}}}& \cdots &0\\
 \vdots &{}& \vdots &{}& \vdots \\
0& \cdots &{{a_{ii}}}&{}&0\\
 \vdots &{}& \vdots &{}& \vdots \\
0& \cdots &{{a_{ni}}}& \cdots &0
\end{array}} \right) = A{E_{ii}} $
For all $i$ with $i \le 1 \le n$. As follows $a_{ij}=0$ für alle $1 \le i,j \le n$ und $i \ne j$.
the second part of the proof: $ \forall 1 \le i \le n$:
$
{E_{1i}}A = \left( {\begin{array}{*{20}{c}}
{{a_{i1}}}& \cdots &{{a_{ii}}}& \cdots &{{a_{in}}}\\
0& \cdots &0& \cdots &0\\
 \vdots & \cdots & \vdots &{}& \vdots \\
 \vdots &{}& \vdots &{}& \vdots \\
0& \cdots &0& \cdots &0
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
0& \cdots &{{a_{11}}}& \cdots &0\\
 \vdots &{}& \vdots &{}& \vdots \\
 \vdots &{}& \vdots &{}& \vdots \\
 \vdots &{}& \vdots &{}& \vdots \\
0& \cdots &{{a_{1n}}}& \cdots &0
\end{array}} \right) = A{E_{1i}}$
so $ a_{ii} = a_{11} \forall 1 \le i \le n \Rightarrow A=aI_n$ for $a \in \mathbb{K}$
Why implies this it must be true for all $ B \in M_{nn}$ ?
 A: Since $AB=BA$ holds $\forall B$, we take an example with $B=E_{ii} $ ($E_{ii}\in M_{nn}(\mathbb{K}))$ which gives us:
$E_{ii}A=AE_{ii}$, and taking all $1\leq i\leq n$, we get $a_{i,j}=0,\, \forall\, i\ne j$
If we take this further using the fact that $A$ is diagonal, taking $B\in M_{nn}(\mathbb{K})$ we have:
$AB= \left[ \begin{array}{ccc}
a_{11} & ...0... & 0 \\
...0... & ... & ...0... \\
0 & ...0... & a_{nn} \end{array} \right]\left[ \begin{array}{ccc}
b_{11} & ... & b_{1n} \\
... & ... & ... \\
b_{n1} & ... & b_{nn} \end{array} \right]=\left[ \begin{array}{ccc}
a_{11}b_{11} & a_{11}b_{12} &... & a_{11}b_{1n} \\
a_{22}b_{21} & ... & ... & ... \\
... & ...& ...&...\\
a_{nn}b_{n1} & ... & ...&a_{nn}b_{nn} \end{array} \right]$, but:
$BA=\left[ \begin{array}{ccc}
b_{11} & ... & b_{1n} \\
... & ... & ... \\
b_{n1} & ... & b_{nn} \end{array} \right]\left[ \begin{array}{ccc}
a_{11} & ...0... & 0 \\
...0... & ... & ...0... \\
0 & ...0... & a_{nn} \end{array} \right]=\left[ \begin{array}{ccc}
a_{11}b_{11} & a_{22}b_{12} &... & a_{nn}b_{1n} \\
a_{11}b_{21} & ... & ... & ... \\
... & ...& ...&...\\
a_{11}b_{n1} & ... & ...&a_{nn}b_{nn} \end{array} \right]$
since the two must be equivalent, comparing the matrices we get $a_{ii}=a_{jj}$ $\forall i,j=1,...,n$ so we get $A=aI$.
