**Definition:** A square matrix $A$ is **diagonalizable** if it is similar to a diagonal matrix, i.e., there exists an invertible matrix $P$ such that:

$$P^{-1}AP = D = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)$$

where $\lambda_i$ are the eigenvalues of $A$.

**Criterion:** $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors (equivalently, the geometric multiplicity equals the algebraic multiplicity for each eigenvalue).

**Procedure:**
1. Find eigenvalues by solving $\det(A - \lambda I) = 0$
2. For each eigenvalue, find eigenvectors by solving $(A - \lambda I)v = 0$
3. If there are $n$ linearly independent eigenvectors, form $P$ from them and $D$ from the eigenvalues

**Example:** $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ has eigenvalues 5 and 2, so $A = PDP^{-1}$ with $D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$.

Diagonalization simplifies computing powers: $A^k = PD^kP^{-1}$.