# $T: V \to V$ is a diagonalizable linear operator. Prove that for any $W\subset V$, there's a $T$-invariant subspace $U$ such that $V=W \oplus U$.

(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 $V=W \oplus U$.

Thanks.

• How would you prove the result if $T$ is the identity operator? Aug 18, 2016 at 21:14
• 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.
– avs
Aug 18, 2016 at 21:15

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?
• 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. Aug 18, 2016 at 21:23
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 $W$?