Eigenvalues and Diagonalization

Diagonalizability and eigenvalue multiplicity

An n\times n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors, equivalently A=PDP^{-1} for some invertible P and diagonal D. For each eigenvalue \lambda, two multiplicities must be distinguished. The algebraic multiplicity m_a(\lambda) is the power of (t-\lambda) in the characteristic polynomial. The geometric multiplicity is m_g(\lambda)=\dim N(A-\lambda I)=n-\operatorname{rank}(A-\lambda I), the number of independent eigenvectors for \lambda. Always 1\le m_g(\lambda)\le m_a(\lambda) for an eigenvalue. The criterion is: A is diagonalizable exactly when m_g(\lambda)=m_a(\lambda) for every eigenvalue \lambda. Over \mathbb{C} the algebraic multiplicities always add up to n, so failure of diagonalizability is always a shortage of eigenvectors.

The main working rules are as follows.

• To find m_g(\lambda), compute \operatorname{rank}(A-\lambda I) on the full matrix and subtract it from n.
• A matrix with n distinct eigenvalues is always diagonalizable, because eigenvectors belonging to distinct eigenvalues are independent. Trouble begins only with repeated eigenvalues.
• For a block upper triangular matrix \begin{pmatrix}B & C\\ 0 & D\end{pmatrix}, the eigenvalues come from the diagonal blocks B and D, but the geometric multiplicities must still be read from \operatorname{rank}(A-\lambda I) of the whole matrix. The sum of geometric multiplicities over the distinct eigenvalues gives the total number of independent eigenvectors.
• Two matrices with the same eigenvalues need not share diagonalizability. What decides it is whether P-\lambda I and Q-\lambda I have the same rank for the repeated eigenvalue. A repeated eigenvalue whose A-\lambda I has rank n-2 gives two eigenvectors and is diagonalizable, whereas rank n-1 on the same multiplicity gives only one and is not.

Common confusions must be avoided. Diagonalizable is not the same as invertible: the matrix J=\begin{pmatrix}2&1\\0&2\end{pmatrix} is invertible with \det J=4, yet J-2I has rank 1, so m_g(2)=1<2=m_a(2) and J is not diagonalizable. Distinct eigenvalues imply diagonalizability, but the converse is false.

Spectral mapping for matrix polynomials

If Av=\lambda v with v\neq 0, then A^k v=\lambda^k v for every k\ge 0. Therefore for any polynomial p, p(A)v=p(\lambda)v. The same eigenvector works for p(A), and p(\lambda) is an eigenvalue of p(A). This is the spectral mapping theorem: the spectrum of p(A) is \{p(\lambda):\lambda \text{ an eigenvalue of } A\}.

The rule is used by substituting each eigenvalue of A into the polynomial, not by substituting into the matrix. If B=A^2+2A+I and p(t)=t^2+2t+1=(t+1)^2, then the eigenvalues of B are p(\lambda)=(\lambda+1)^2 for each eigenvalue \lambda of A. Terms such as 2A and I must be included in p; leaving them out gives the wrong values. Distinct eigenvalues of A can merge under p: for p(t)=t^2, the eigenvalues 1 and -1 of A both give 1 for p(A).