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Let $\mathbb{K}$ a field, $V$ a vector space over $\mathbb{K}$. If $T:V\to V$ commutes with all other linear operators $V \to V$, then there exists $\lambda \in \mathbb{K}$ such that $T= \lambda I$, where $I$ is the identity V.

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marked as duplicate by Marc van Leeuwen, Shobhit, Lord_Farin, Dennis Gulko, Stefan Hansen Oct 30 '13 at 13:07

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First show that $Tv$ is scalar multiple of $v$ for all $v\in V$. You can do this by assuming, for the sake of contradiction, that there exists $v$ such that the vectors $v$ and $Tv=w$ are linearly independent. Form a basis $\mathcal B$ which includes both $v$ and $w$. Now, define a new linear operator $L:V\to V$ which maps every basis element to $0$ except $w$, which we map to $v$. You will get a contradiction.

Now, you must show that there is just one scalar $\lambda$ such that $Tv=\lambda v$ for all $v$. Say $Tv_1=\lambda_1v_1$ and $Tv_2=\lambda_2v_2$ where neither $v_1$ nor $v_2$ is a scalar multiple of the other--i.e. they are linearly indepdendent--notice that the case where they are linearly dependent is easily handled. Write $T(v_1+v_2)=\lambda_3(v_1+v_2)$. Then the equation $$(\lambda_3-\lambda_1)v_1+(\lambda_3-\lambda_2)v_2=0$$ implies $\lambda_1=\lambda_3=\lambda_2$ as desired.

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Hint What matrix Commutes With Every Other matrix? in general what is the center of $M_n(\mathbb{R})$

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  • $\begingroup$ This answer only works for finite dimensional $V$. $\endgroup$ – Karl Kronenfeld Oct 30 '13 at 12:06

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