Diagonalization

Choosing the Right Coordinates

Eigenvectors give a matrix its "natural axes." Diagonalization exploits this: if a matrix has enough independent eigenvectors to form a basis, then in that basis the transformation is simply a scaling along each axis — a diagonal matrix. Complicated becomes simple, just by choosing the right coordinates.

Concretely, a square matrix $A$ is diagonalizable if it can be written $$A = P D P^{-1},$$ where $D$ is diagonal (the eigenvalues on its diagonal) and $P$ has the corresponding eigenvectors as its columns. The condition is exactly having a basis of eigenvectors: an $n \times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors. (A sufficient condition: $n$ distinct eigenvalues guarantees it, since eigenvectors for distinct eigenvalues are automatically independent.)

Not every matrix is diagonalizable — a matrix with a repeated eigenvalue that lacks enough independent eigenvectors (a defective matrix) cannot be diagonalized. So diagonalizability is a genuine property, not automatic.

The payoff is computational. Powers become trivial: $A^k = P D^k P^{-1}$, and $D^k$ just raises each diagonal entry to the $k$. Computing $A^{100}$ by repeated multiplication is enormous work; via diagonalization it is one matrix of easy powers, sandwiched by $P$ and $P^{-1}$. This is exactly how one solves systems of linear recurrences and differential equations, and how the long-term behavior of dynamical systems (dominant eigenvalue) is read off. Diagonalization is the practical reason eigenvalues matter.

Common pitfall: assuming every square matrix is diagonalizable. Diagonalization requires a full set of $n$ independent eigenvectors; a defective matrix (too few independent eigenvectors, from a repeated eigenvalue) cannot be diagonalized. Having $n$ distinct eigenvalues is sufficient but not necessary — some repeated-eigenvalue matrices are still diagonalizable, and some are not; you must check the eigenvectors.

A skewed eigenvector grid straightened by P^{-1} into axis-aligned coordinates where the accent diagonal matrix D simply scales each axis, then mapped back by P — diagonalization as a change of coordinates.

$$A = P D P^{-1},\quad A^k = P D^k P^{-1}$$