If \(T\) is a diagonalizable linear operator, then every \(T\)-invariant subspace has a complementary \(T\)-invariant subspace.
Solution
Let the distinct eigenvalues of \(T\) be \(\lambda_1,\ldots,\lambda_r\), with eigenspaces \[ E_i=\ker(T-\lambda_iI). \] Diagonalizability gives \[ V=E_1\oplus\cdots\oplus E_r \] [@HK71]. Let \(W\subseteq V\) be \(T\)-invariant.
1. The subspace \(W\) splits across the eigenspaces.
For each \(i\), the Lagrange polynomial \[ p_i(x)=\prod_{j\ne i}\frac{x-\lambda_j}{\lambda_i-\lambda_j} \] defines the spectral projection \(p_i(T)\) onto \(E_i\). Since \(W\) is invariant under \(T\), it is invariant under every polynomial in \(T\). Therefore \[ p_i(T)W\subseteq W. \] If \(w=\sum_iw_i\) is the eigenspace decomposition of \(w\in W\), then \(w_i=p_i(T)w\in W\). Hence \[ W=(W\cap E_1)\oplus\cdots\oplus(W\cap E_r). \]
2. Choose complements inside the eigenspaces.
For each \(i\), choose an ordinary vector-space complement \(F_i\) such that \[ E_i=(W\cap E_i)\oplus F_i. \] This is possible by extending a basis of \(W\cap E_i\) to a basis of \(E_i\). Every subspace of \(E_i\) is \(T\)-invariant, because \(T\) acts on \(E_i\) as the scalar \(\lambda_i\). Thus each \(F_i\) is \(T\)-invariant.
3. Their direct sum is the required invariant complement.
2. Combining the decompositions in steps<1>1–<1>2 with \(V=\bigoplus_iE_i\) gives\[ V=W\oplus F. \]Hence every invariant subspace of a diagonalizable operator has acomplementary invariant subspace.