Diagonalization is the ideal form of a linear operator: the operator acts independently on coordinate axes. Triangularization is weaker but far more common, and it is often enough for computing powers, determinants, traces, and characteristic polynomials.
Diagonalization
Definition. An operator \(T:V\to V\) is diagonalizable over \(F\) if there exists a basis of \(V\) made of eigenvectors of \(T\). Equivalently, some matrix of \(T\) is diagonal. A matrix \(A\) is diagonalizable if \(A=PDP^{-1}\) for an invertible \(P\) and diagonal \(D\).
Theorem. A finite-dimensional operator \(T\) is diagonalizable if and only if \[ V=\bigoplus_{\lambda}E_\lambda, \] the direct sum of its eigenspaces. Equivalently, the sum of the geometric multiplicities of all eigenvalues equals \(\dim V\).
Proof. A basis of eigenvectors is the same as choosing bases inside eigenspaces whose union is a basis of \(V\). Eigenspaces for distinct eigenvalues intersect trivially and their vectors are independent, so their sum is direct. The dimension condition says exactly that the direct sum fills \(V\).
Two practical criteria
Proposition. If \(\chi_T\) splits into distinct linear factors, then \(T\) is diagonalizable. More generally, \(T\) is diagonalizable over \(F\) if and only if its minimal polynomial splits into distinct linear factors.
Proof. Distinct eigenvalues provide independent eigenspaces, and if there are \(n\) distinct roots in dimension \(n\), the corresponding eigenvectors already form a basis. For the minimal polynomial criterion, the squarefree polynomial splits into coprime factors \((t- \lambda)\); Bezout identities for these factors produce projections onto the eigenspaces. Repeated factors are exactly the obstruction created by nontrivial nilpotent parts.
Triangularization
Theorem. If the characteristic polynomial of \(T\) splits over \(F\), then \(T\) has an upper triangular matrix in some basis. In particular, every complex matrix is triangularizable.
Proof. Since \(\chi_T\) splits, \(T\) has an eigenvector \(v_1\). Extend \(v_1\) to a basis; the first column has zeros below the first entry. The induced operator on \(V/\operatorname{span}(v_1)\) has characteristic polynomial \(\chi_T/(t- \lambda_1)\), which also splits. Induct on dimension.
Information visible from triangular form
Proposition. If \(A\) is triangular with diagonal entries \(\lambda_1, \ldots, \lambda_n\), then \[ \det A=\prod_i\lambda_i, \qquad \operatorname{tr}A=\sum_i\lambda_i, \qquad \chi_A(t)=\prod_i(t- \lambda_i). \] Thus determinant and trace are sums and products of eigenvalues counted with algebraic multiplicity whenever \(\chi_A\) splits.
Non-diagonalizable matrices
Counterexample. The matrix \[ A=\begin{pmatrix}1&1\\0&1\end{pmatrix} \] has only the eigenvalue \(1\). Its eigenspace is \(\ker(A-I)=\operatorname{span}(e_1)\), one-dimensional, so it cannot have an eigenbasis. It is triangular but not diagonalizable.
Commuting diagonalizable operators
Proposition. If two diagonalizable operators \(S,T\) commute and their minimal polynomials split, then they can be simultaneously diagonalized. Each eigenspace of \(S\) is \(T\)-invariant, and diagonalizing \(T\) inside those eigenspaces gives a common eigenbasis.
Repeated eigenvalues and missing eigenvectors
Example. The identity matrix \(I_2\) has characteristic polynomial \((t-1)^2\) and two independent eigenvectors. The Jordan block \[ J=\begin{pmatrix}1&1\\0&1\end{pmatrix} \] has the same characteristic polynomial, but \(J-I\) has kernel \(\operatorname{span}\{e_1\}\), so there is only one independent eigenvector. Therefore \(I_2\) is diagonalizable and \(J\) is not.
Counterexample. Having all eigenvalues in the field is not enough for diagonalization. The obstruction is not the characteristic polynomial alone; it is the size of the eigenspaces. One must compare algebraic multiplicity with geometric multiplicity.