Show that a linear map \(T:V\to V\) is diagonalizable if and only if \[ V=\bigoplus_\lambda \ker(T-\lambda I), \] where the sum runs over the eigenvalues of \(T\) in the base field.
Solution
If \(T\) is diagonalizable, choose a basis of eigenvectors and group the basis vectors according to their eigenvalues. The span of the vectors with eigenvalue \(\lambda\) is exactly \[ E_\lambda=\ker(T-\lambda I), \] and the basis partition shows \[ V=\bigoplus_\lambda E_\lambda. \]
Conversely, suppose \[ V=\bigoplus_\lambda E_\lambda. \] Choose a basis of each eigenspace \(E_\lambda\). The union of these bases is a basis of \(V\), and every vector in it is an eigenvector of \(T\). Hence the matrix of \(T\) in this basis is diagonal.
Thus \(T\) is diagonalizable exactly when \(V\) is the direct sum of its eigenspaces.