Diagonalization and the Spectral Theorem

Suppose an $n \times n$ matrix $A$ has $n$ independent eigenvectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ with eigenvalues $\lambda_1, \dots, \lambda_n$. Put the eigenvectors in the columns of $P$ and the eigenvalues on the diagonal of $\Lambda$. The $n$ equations $A\mathbf{v}_i = \lambda_i\mathbf{v}_i$ stack into one matrix equation $AP = P\Lambda$, and because the columns of $P$ are independent, $P$ is invertible:

This is diagonalization. Read it right to left as three moves: $P^{-1}$ rewrites a vector in eigenvector coordinates, $\Lambda$ scales each coordinate by its own eigenvalue, and $P$ translates back. In the eigenvector basis the matrix is a list of separate scalings.

Example: $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ has trace 7 and determinant 10, so $\lambda^2 - 7\lambda + 10 = 0$ gives $\lambda = 5, 2$. Solving $(A - 5I)\mathbf{v} = \mathbf{0}$ gives $(1, 1)$, and $(A - 2I)\mathbf{v} = \mathbf{0}$ gives $(1, -2)$. So

When does it fail? Eigenvectors for distinct eigenvalues are independent, so any matrix with $n$ distinct eigenvalues is diagonalizable. Failure needs a repeated eigenvalue with too few eigenvectors. The shear $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has $\lambda = 1$ twice, but $A - I$ has rank 1, so its only eigenvectors are multiples of $(1, 0)$. Such a matrix is called defective. A repeated eigenvalue on its own is harmless: $2I$ has $\lambda = 2$ twice and is already diagonal.