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Prove or give a counterexample: given a vector space $V$ over the complex numbers and $T$ a linear operator. If $T$ is a projection ($T^2 =T$), then $T$ is diagonalizable. I know this is true if $V$ is finite-dimensional because then $T$ is completely reducible and the minimal polynomial factors into linear factors. But is this statement also true when $V$ is infinite-dimensional?

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    $\begingroup$ what do you understand under diagonalizability when dealing with infinite dimensional vector spaces? $\endgroup$ – user190080 Nov 21 '15 at 20:43
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This is always true. Let $K=\ker(T)$ and $I=\operatorname{im}(T)$. Since $T^2=T$, $K\cap I=0$, and furthermore for any $v\in V$, $v=Tv+(1-T)v\in I+K$. So $V$ is the direct sum of the subspaces $K$ and $I$. Now choose any basis of $K$ and any basis of $I$; the union of these two bases will then be a basis of $V$ which diagonalizes $T$.

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