# Proving a diagonal matrix exists for linear operators with invariant subspaces

I came across this problem one of my practice worksheets and I was stumped as to how I would go about solving this.

Let $T : V \rightarrow V$ be a linear operator on a finite dimensional vector space $V$ over $C$. Assume that $T$ has the following property: for each invariant subspace $U \subset V$ , there exists an invariant subspace $W \subset V$ such that $V = U \oplus W$. Show that $T$ has a diagonal matrix with respect to some basis of $V$.

One thing I know is that as a direct result of the last assumption, that we have dim V linearly independent vectors that make up the bases of $V$ and $U$, but I'm not sure how exactly to go from here to showing that these vectors are eigenvectors of T.

Any help would be greatly appreciated. Thanks!

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Proceed by induction on $\dim V$. If $\dim V=1$ the result is trivial. Suppose $\dim V>1$. By the Fundamental Theorem of Algebra, $\mathrm{char}_T(z)$ has some root in $\mathbb C$, say $\lambda$. Thus there is some eigenvector $v$ of $T$ with eigenvalue $\lambda$. Let $U=\mathrm{span}\{v\}$, which is invariant under $T$ since $v$ is an eigenvector of $T$, and let $W$ be a subspace invariant under $T$ such that $V=U\oplus W$. If we choose a basis $b_1,\ldots,b_n$ for $V$ such that $b_1=v$ and $b_2,\ldots,b_n\in W$, then in this basis we have $$T=\begin{pmatrix} T|_U & 0 \\ 0 & T|_W\end{pmatrix}=\begin{pmatrix}\lambda & 0\\ 0 & T|_W\end{pmatrix}$$ and since $\dim W<\dim V$, by the inductive hypothesis we can choose $b_2,\ldots,b_n$ such that $T|_W$ is diagonal, thus $T$ is diagonal.