Let \(V\) be a finite-dimensional vector space over a field \(K\), and let \(T: V \to V\) be a linear operator. Show that \(T\) is diagonalizable if and only if \(V\) is the direct sum of its eigenspaces: \[V = \bigoplus_{i=1}^k E_{\lambda_i}(T) = \bigoplus_{i=1}^k \ker(T - \lambda_i I)\] where \(\lambda_1, \dots, \lambda_k \in K\) are the distinct eigenvalues of \(T\).
Solution
For each eigenvalue \(\lambda\) of \(T\), write \[ E_\lambda=\ker(T-\lambda I). \] Eigenspaces for distinct eigenvalues are linearly independent: if \[ v_1+\cdots+v_r=0,\qquad v_i\in E_{\lambda_i}, \] with the \(\lambda_i\) distinct, then applying \[ \prod_{j\ne i}(T-\lambda_jI) \] gives \[ \left(\prod_{j\ne i}(\lambda_i-\lambda_j)\right)v_i=0, \] so \(v_i=0\). Hence the sum of the eigenspaces is always direct.
If \(T\) is diagonalizable, \(V\) has a basis of eigenvectors. Every basis vector lies in one of the eigenspaces, so the eigenspaces span \(V\). Therefore \[ V=\bigoplus_\lambda E_\lambda. \]
Conversely, suppose \[ V=\bigoplus_\lambda E_\lambda. \] Choose a basis of each eigenspace and take their union. Because the sum is direct and equals \(V\), this union is a basis of \(V\), and every one of its vectors is an eigenvector of \(T\). Thus the matrix of \(T\) in this basis is diagonal, so \(T\) is diagonalizable.
Equivalently, \[ T\text{ is diagonalizable} \iff \sum_\lambda \dim E_\lambda=\dim V. \]