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(Couldn't think of a better shorter title)

Let $T: V \to V$ be a diagonalizable linear operator. Prove that for any subspace $W$ of $V$, there is a subspace $U$ that is invariant under $T$ such that $V=W \oplus U$.

I look for a hint, something to start with. Please do not supply a full proof.

Thanks.

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  • $\begingroup$ How would you prove the result if $T$ is the identity operator? $\endgroup$
    – anomaly
    Aug 18, 2016 at 21:14
  • $\begingroup$ Let's start by just writing down the invariant definition of "diagonalizable": it means that there exist 1-dimensional subspaces $U_{1}, U_{2}, \ldots, U_{N}$ that are each $T$-invariant and such that $V$ equals their direct sum. $\endgroup$
    – avs
    Aug 18, 2016 at 21:15

2 Answers 2

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Suppose $\dim V = n > 0$. Choose a basis $\{v_1, \ldots, v_k\}$ of $W$. Extend this to a basis $\{v_1, \ldots, v_k, v_{k+1}, \ldots, v_n\}$ of $V$ consisting of eigenvectors of $T$. Then, what allows us to conclude that $W \oplus \operatorname{span}(v_{k+1},\ldots,v_n)$ is a direct sum?

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  • $\begingroup$ Why does $W$ have a basis consisting of eigenvectors of $T$? The subspace $W$ needs to be $T$-invariant in order for the restriction $T \vert_{W}$ to be diagonalizable. Regardless, you can go through the same proof by starting with an ordinary basis of $W$, and then completing to a basis of $V$ where the vectors $v_{k+1}, \ldots, v_{n}$ are eigenvectors of $T$. Although, I think the OP only asked for a hint, and this answer says maybe too much. $\endgroup$ Aug 18, 2016 at 21:23
  • $\begingroup$ @AlexWertheim, noted. $\endgroup$
    – Alex Ortiz
    Aug 18, 2016 at 21:25
  • $\begingroup$ It is maybe too much, but I was already thinking along this line, so still ok... :) The simple question in the end actually pointed me into the right direction. So thanks. $\endgroup$
    – Yes
    Aug 18, 2016 at 22:05
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You can try by recursion. Start with $\dim W = \dim V -1$. Let $(e_1, \cdots, e_n)$ be a basis of $V$ which diagonalizes $T$. Is it possible that $(e_1, \cdots, e_n)$ is included in $W$?

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