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Prove that the only $3×3$ matrices which commute with any $3×3$ matrices are of the form $cI$ for some scalar $c$.

-My professor's hint for me was that if $A$ is such a matrix. That by choosing $B$ wisely, comparing $AB$ to $BA$ should narrow the choices for the entries of $A$ fairly quickly. -I understand that I could find a particular matrix for which this condition would fit, although I do not see how I could prove this for all A matrices, any help is appreciated.

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Let $A:=(a_{i,j})_{1\leqslant i,j\leqslant n}\in\mathbb{R}^{n\times n}$ and assume that: $$\forall B\in\mathbb{R}^{n\times n},AB=BA.$$ For all $(i,j)\in\{1,\ldots,n\}^2$ let define: $$E_{i,j}:=(\delta_{i,k}\delta_{j,l})_{1\leqslant k,l\leqslant n}\in\mathbb{R}^{n\times n}.$$ $E_{i,j}$ is a square matrix of order $n$ with $0$ everywhere except on the entry $(i,j)$ where it has a $1$. Notice that for all $(i,j)\in\{1,\ldots,n\}^2$, one has: $$AE_{i,j}=\begin{pmatrix}0 & \ldots & 0 & a_{1,i} & 0 & \ldots & 0\\ \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & \ldots & 0 & {a_{n,i}} & 0 & \ldots & 0\end{pmatrix},$$ where the entries of $A$ are on the $j$th column. Moreover, one has: $$E_{i,j}A=\begin{pmatrix}0 & \ldots & 0\\ \vdots & \ddots & \vdots\\ 0 & \ldots & 0\\ a_{j,1} & \ldots & a_{j,n}\\ 0 & \ldots & 0\\ \vdots & \ddots & \vdots\\ 0 & \ldots & 0\\ \end{pmatrix},$$ where the entries of $A$ are on the $i$th row. Assuming $i\neq j$, since $AE_{i,j}=E_{i,j}A$ (by hypothesis), by checking the entry $(j,j)$, one has: $$a_{j,i}=0.$$ From there, the entries of $A$ are $0$ everywhere except on its diagonal. I let you conclude from there.

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  • $\begingroup$ Just to confirm.... to conclude.... Since there is only entries on the diagonal it is $I$ or some variation $cI$ as the problem stated. Thus concluding the proof? $\endgroup$ – Pablo Sanchez Nov 14 '15 at 22:32
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    $\begingroup$ There is still one step left in the proof, you have to show that all the entries on the diagonal are equaled. $\endgroup$ – C. Falcon Nov 14 '15 at 22:35
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    $\begingroup$ +1. Just note that to me it seems that $\mathcal{M}_n(\mathbb{R})$ is quite a French notation, the internationally recognized notation is $\mathbb{R}^{n\times n}$. $\endgroup$ – yo' Nov 14 '15 at 22:38
  • $\begingroup$ You're certainly right, so I fixed it! Thank you! :) $\endgroup$ – C. Falcon Nov 14 '15 at 22:40
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    $\begingroup$ @PabloSanchez If $a_{i,i}\neq a_{j,j}$, then $A$ does not commute with the matrix that has all zeros but an element in position $(i,j)$. $\endgroup$ – yo' Nov 14 '15 at 22:55

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