Let \(T\) be a diagonalizable linear operator on the \(n\)-dimensional vector space \(V\), and let \(W\) be a subspace which is invariant under \(T\). Prove that the restriction operator \(T_W\) 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, \[ V=E_1\oplus\cdots\oplus E_r \] [@HK71].
1. Each spectral projection is a polynomial in \(T\) and preserves \(W\).
For each \(i\), define the Lagrange polynomial \[ p_i(t)=\prod_{j\ne i}\frac{t-\lambda_j}{\lambda_i-\lambda_j}. \] The denominators are nonzero because the eigenvalues are distinct. For \(v\in E_k\), \[ p_i(T)v=p_i(\lambda_k)v =\begin{cases}v,&k=i,\\0,&k\ne i.\end{cases} \] Thus \(p_i(T)\) is precisely the projection of \(V\) onto \(E_i\) along the direct sum of the other eigenspaces.
Because \(W\) is \(T\)-invariant, it is invariant under every power \(T^m\), hence under every polynomial in \(T\). Therefore \[ p_i(T)(W)\subseteq W \] for every \(i\).
2. The invariant subspace is the direct sum of its intersections with the eigenspaces.
Take \(w\in W\). Its unique eigenspace decomposition is \[ w=w_1+\cdots+w_r, \qquad w_i\in E_i. \] By step 1, \[ w_i=p_i(T)w\in W. \] Hence \(w_i\in W\cap E_i\) for every \(i\), so \[ W\subseteq\sum_{i=1}^r(W\cap E_i). \] The reverse inclusion is immediate, and the sum is direct because the \(E_i\) already form a direct sum. Therefore \[ W=(W\cap E_1)\oplus\cdots\oplus(W\cap E_r). \]
3. A basis adapted to this decomposition diagonalizes \(T_W\).
Choose a basis of each subspace \(W\cap E_i\) and concatenate these bases. Step 2 shows that the resulting list is a basis of \(W\). Every vector in the basis belonging to \(W\cap E_i\) satisfies \[ T_Wv=Tv=\lambda_i v. \] Thus this is a basis of \(W\) consisting entirely of eigenvectors of \(T_W\). Consequently \(T_W\) is diagonalizable.