MATH319 Slides 25 Functions of a matrix 27 Polynomials of a diagonable matrix

26 Characteristic equation

The eigenvalue equation is

Av=\lambda v

where $v\neq 0$ is the eigenvector and $\lambda\in{\bf C}$ the eigenvalue.

Lemma

The eigenvalues of $n\times n$ complex matrix $A$ are the roots of the characteristic equation

\chi_{A}(\lambda)=0.

Proof. Recall that $\det(sI-A)=\chi_{A}(s)$ . By the Fundamental Theorem of Algebra, there are $n$ complex roots, counted according to algebraic multiplicity. When $\chi_{A}(\lambda)=0,$ the matrix $\lambda I-A$ is not invertible, so there exists $v\in V$ , with $v\neq 0$ and $Av=\lambda v$ and $\lambda$ is an eigenvalue. Conversely, if there exists $v\neq 0$ and $\lambda\in{\bf C}$ such that $Av=\lambda v$ , then $\lambda I-A$ is not invertible, so $\chi_{A}(\lambda)=0$ .