Let \(V\) be finite-dimensional and let \(T:V\to V\) be diagonalizable. If \(W\subseteq V\) is a linear subspace with \(T(W)\subseteq W\), prove that the restriction \(T|_W\) is diagonalizable.
Solution
Let \[ \lambda_1,\ldots,\lambda_r \] be the distinct eigenvalues of \(T\), and let \[ E_i=\ker(T-\lambda_iI). \] Since \(T\) is diagonalizable, \[ V=E_1\oplus\cdots\oplus E_r. \]
1. The subspace \(W\) is invariant under every polynomial in \(T\).
The assumption \[ T(W)\subseteq W \] implies inductively that \[ T^k(W)\subseteq W \] for every \(k\geq0\). Hence for any polynomial \[ p(t)=a_0+a_1t+\cdots+a_mt^m, \] one has \[ p(T)(W) \subseteq W. \]
2. For each \(i\), define\[ p_i(t) = \prod_{j\neq i} \frac{t-\lambda_j}{\lambda_i-\lambda_j}. \]Then \(p_i(T)\) is the projection of \(V\) onto \(E_i\) along the sum of the other eigenspaces.
The Lagrange polynomial \(p_i\) satisfies \[ p_i(\lambda_i)=1 \] and \[ p_i(\lambda_j)=0 \qquad (j\neq i). \] If \(v\in E_j\), then \[ T(v)=\lambda_jv, \] so \[ p_i(T)v=p_i(\lambda_j)v. \] Thus \(p_i(T)\) acts as the identity on \(E_i\) and as zero on every \(E_j\) with \(j\neq i\). Since the \(E_j\) form a direct-sum decomposition of \(V\), this is exactly the stated projection.
3. If\[ w=w_1+\cdots+w_r \]is the eigenspace decomposition of a vector \(w\in W\), with \(w_i\in E_i\), then every \(w_i\) belongs to \(W\).
By step 2, \[ w_i=p_i(T)w. \] By step 1, \(p_i(T)\) preserves \(W\). Since \(w\in W\), it follows that \[ w_i\in W. \]
4. One has\[ W = \bigoplus_{i=1}^r(W\cap E_i). \]
Every vector \(w\in W\) has an eigenspace decomposition in \(V\). Step 3 shows that each component lies in the corresponding intersection \(W\cap E_i\), so \[ W\subseteq\sum_i(W\cap E_i). \] The reverse inclusion is immediate. The sum is direct because the sum of the \(E_i\) in \(V\) is direct.
5. The restriction \(T|_W\) is diagonalizable.
For each \(i\), every vector in \(W\cap E_i\) is an eigenvector of \(T|_W\) with eigenvalue \(\lambda_i\). By step 4, the direct sum of these eigenspaces is all of \(W\). Choosing a basis of each nonzero \(W\cap E_i\) and taking their union gives a basis of \(W\) consisting of eigenvectors of \(T|_W\). Hence \(T|_W\) is diagonalizable.
6. Q.E.D.
Step 5 is the required conclusion.