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This question has been answered here: https://math.stackexchange.com/a/181768

But I am confused on the last line, when the author writes:

Well, compare the different cases and deduce $c = a_{ji}$

How exactly is this deduction done in more detail? In particular, while I can see that $T(v_i) = a_{ji} v_i$, I'm not sure how to conclude that $a_{ji} = c$ for all $i, j$ which is needed to finish the proof.

A similar proof is found at A linear operator commuting with all such operators is a scalar multiple of the identity., but that one uses a different proof strategy - I am more interested in finding out how to finish up this particular proof, not simply giving any proof of the statement.

greg115
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    That answer has two errors. The first is perhaps a typo (the evaluation of $TS_j(v_i)$) but the second is a logical error: Using that set of linear maps ${S_i }$ only shows $T$ is a diagonal matrix. Specifically, you can show $a_{ji}=0$ for $j\neq i$ but to show $a_{ii}=a_{jj}$ you need to use another type of "$S$" – Brian Moehring Oct 05 '22 at 03:45

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I took a look at the answers. I prefer the following explanation.

Assume $y:=Tx$ and $x$ are linearly independent. There is a linear transformation $S$ so that $Sx=x$ and $Sy=2y.$ Then $STx=2y$ and $TSx=y, $ a contradiction. Thus $Tx=\lambda_x x$ for any $x.$ Assume $\lambda_y\neq \lambda_x$ for some $x,y\neq 0.$ Then $x$ and $y$ are linearly independent. There is $S$ such that $Sx=y,\ Sy=x.$ Then $STx=\lambda_xy$ and $TSx=\lambda_yy,$ a contradiction. Hence $T=\lambda I.$

Ryszard Szwarc
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