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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:

$ % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaGaamyramaaBaaaleaacaWGPbGaamyAaa % qabaGccaWGbbGaeyypa0ZaaeWaaeaafaWabeqbfaaaaaqaaiaaicda % aeaacqWIVlctaeaacaaIWaaabaGaeS47IWeabaGaaGimaaqaaiabl6 % Uinbqaaaqaaiabl6Uinbqaaaqaaiabl6UinbqaaiaadggadaWgaaWc % baGaamyAaiaaigdaaeqaaaGcbaGaeS47IWeabaGaamyyamaaBaaale % aacaWGPbGaamyAaaqabaaakeaaaeaacaWGHbWaaSbaaSqaaiaadMga % caWGUbaabeaaaOqaaiabl6Uinbqaaaqaaiabl6Uinbqaaaqaaiabl6 % UinbqaaiaaicdaaeaacqWIVlctaeaacaaIWaaabaGaeS47IWeabaGa % aGimaaaaaiaawIcacaGLPaaacqGH9aqpdaqadaqaauaadeqafuaaaa % aabaGaaGimaaqaaiabl+UimbqaaiaadggadaWgaaWcbaGaaGymaiaa % dMgaaeqaaaGcbaGaeS47IWeabaGaaGimaaqaaiabl6Uinbqaaaqaai % abl6Uinbqaaaqaaiabl6UinbqaaiaaicdaaeaacqWIVlctaeaacaWG % HbWaaSbaaSqaaiaadMgacaWGPbaabeaaaOqaaaqaaiaaicdaaeaacq % WIUlstaeaaaeaacqWIUlstaeaaaeaacqWIUlstaeaacaaIWaaabaGa % eS47IWeabaGaamyyamaaBaaaleaacaWGUbGaamyAaaqabaaakeaacq % WIVlctaeaacaaIWaaaaaGaayjkaiaawMcaaiabg2da9iaadgeacaWG % fbWaaSbaaSqaaiaadMgacaWGPbaabeaaaaa!8071! {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}$ ?

fast-forward
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1 Answers1

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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 $A=aI$.

Ellya
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  • but it is just a example why implies this example that it is true $ \forall B $ if $ A = aI_n $ ? – fast-forward Apr 09 '14 at 19:29
  • it was the first step, see my edit. – Ellya Apr 09 '14 at 20:04
  • As i understand you've shown elegantly that $ a_{ii} \forall i=1,...,n $ must have the same value to show that the two matrices are identical and this implies $A=aI_n$. But I don't understand why the author of my given solution shows that if it is true for $E_{ii}$ it must be true for all $B$ – fast-forward Apr 09 '14 at 20:46
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    I see! He means it the other way. Because it is true for all $B$ it is true for $E_{ii}$. – Ellya Apr 09 '14 at 21:02
  • While your proof appears to me complete in the other case I ask myself: if the condition is that it is true for all $B$ why it is sufficient to take two types of matrices from the set and show that it can only be true if $A=I_n$. But probably it was not the task to show that it is true for all B. But for me your variant is significantly simpler and more elegant. – fast-forward Apr 09 '14 at 21:39
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    Basically it was to prove that if it holds for all B, it must be that A=aI, so nicely in this case, taking a few different cases of "B", gives us the correct restrictions on A. – Ellya Apr 09 '14 at 21:41
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    Also, if $A=aI$, then $AB=aIB=aB$ and $BA=BaI=Ba$, since we can commute with a scalar we get $AB=BA,,\forall B$ – Ellya Apr 09 '14 at 21:46
  • this was the solution i've given in an exercise but i've only got 2 instead of 10 points. it was commented: "trivial!?" and "show that only $A=aI_n$ could be the solution" – fast-forward Apr 09 '14 at 22:19
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    Its literally because they wanted you to show the other way :( – Ellya Apr 09 '14 at 22:21
  • Well, maybe you can not always know what they want ;) – fast-forward Apr 09 '14 at 22:42