TITLE: Diagonalization of restrictions of a diagonalizable linear operator
QUESTION [9 upvotes]: I realized that I have some difficulties for prove this exercise.
Let $T : V \rightarrow V$ be a linear operator on a finite dimensional vector space $V$ over a field $F$, and invariant subspaces $U,W \subset V$  such that $V = U \oplus W$. Show that if $T$ is diagonalizable then $T_{|U}, T_{|W}$ are  diagonalizable.
Any help would be greatly appreciated. Thanks!

REPLY [8 votes]: There is an elementary proof of the more general statement that the restriction of a diagonalisable linear operator$~T$ to a $T$-stable subspace $U$ is again diagonalisable (in the finite dimensional case), along the lines of my other answer. See also this answer for a variant formulation.
Any vector of $u\in U$ decomposes uniquely in$~V$ as $u=v_1+\cdots+v_k$, a sum of eigenvectors for distinct eigenvalues $\lambda_1,\ldots,\lambda_k$, and it suffices to show that those eigenvectors $v_i$ lie in$~U$. Since $U$ is $T$-stable, it is also stable under any $T-\lambda I$. Now to show that $v_i\in U$, apply to the equation successively $T-\lambda_j I$ for $j\in\{1,2,\ldots,k\}\setminus\{i\}$. At each application each term $v_j$ is multiplied by a scalar, which is zero if and only if $T-\lambda_j I$ was being applied. The result is that only a nonzero scalar multiple of $v_i$ remains, and since we started out with a vector $u$ of $U$, this result is still in$~U$. After division by the scalar this shows that $v_i\in U$.
An equivalent formulation is by induction on$~k$, the number of (nonzero!) eigenvectors needed in the sum for$~u$. The starting cases $k\leq1$ are obvious. Otherwise application of $T-\lambda_kI$ gives
$$
 Tu-\lambda_ku=(\lambda_1-\lambda_k)v_1+(\lambda_2-\lambda_k)v_2+\cdots+(\lambda_{k-1}-\lambda_k)v_{k-1},
$$
and one can apply the induction hypothesis to the vector $Tu-\lambda_ku\in U$ to conclude that all the individual terms (eigenvectors) in the right hand side lie in $U$. But then so do the unscaled $v_1,\ldots,v_{k-1}$ (since all scalar factors are nonzero), and by necessity the remaining term $v_k$ in $u=v_1+\cdots+v_k$ must also lie in$~U$.