Diagonalization means an eigenbasis
An operator is diagonalizable when there's a basis of the whole space made entirely of eigenvectors — an eigenbasis. In that basis A acts by independent scalings, so it becomes the diagonal matrix of eigenvalues. Concretely A = P D P^-1, where the columns of P are the eigenvectors and D is diagonal. The whole game is: can you assemble n independent eigenvectors?
So diagonalizability is a property of the operator, not a calculation you perform — either an eigenbasis exists or it doesn't, regardless of how cleverly you compute. Reframe the question as a counting problem: each eigenvalue contributes the dimension of its eigenspace, and you need those dimensions to add up to n. The rest of this guide makes that count exact.
Independence across eigenvalues
A key lemma: eigenvectors belonging to distinct eigenvalues are automatically linearly independent. So eigenspaces for different lambdas overlap only at the zero vector — their sum is direct. This means you can build a global candidate basis by simply concatenating a basis of each eigenspace; no eigenvector from one space can be a hidden combination of others.
The criterion, stated cleanly
Now the full diagonalizability criterion: an operator over an algebraically closed field is diagonalizable iff geometric multiplicity equals algebraic multiplicity for every eigenvalue. Equivalently, the eigenspace dimensions sum to n — you have enough eigenvectors to fill a basis. The distinct-eigenvalue shortcut is just the case where every multiplicity is 1.
- Find all eigenvalues and their algebraic multiplicities by factoring the characteristic polynomial.
- For each lambda compute dim E_lambda; if it equals the algebraic multiplicity for every eigenvalue, A is diagonalizable.
- Concatenate eigenspace bases to form P, place eigenvalues in D, and verify A = P D P^-1 — which makes powers A^k = P D^k P^-1 trivial.