EIGENVALUES and EIGENVECTORS

Definitions

Given an $n\times n$ matrix A, some $\lambda$ and ${\bf x}\neq {\bf0}$, with

$$A{\bf x}= \lambda{\bf x}.$$

• $\lambda$ is an eigenvalue of A.
• ${\bf x}$ is an eigenvector of A corresponding to $\lambda$.
• The $\bf eigenspace$ for some $\lambda$ is $\{{\bf x}: A{\bf x}=\lambda{\bf x}\}$.

Properties

• $\lambda$ is an eigenvalue of A iff the equation $(A-\lambda I ){\bf x}= {\bf0}$ has a nontrivial solution.
• Diagonal entries of triangular matrices are eigenvalues.
• A is invertible iff 0 is not an eigenvalue for A.
• Eigenvectors for distinct eigenvalues are independent.

The CHARACTERISTIC EQUATION

Characteristic Polynomial

• $\lambda$ is an eigenvalue for A iff $det(A-\lambda I) = 0$.
• The roots of characteristic polynomial
  $p(\lambda) = det(A-\lambda I)$, are eigenvalues for A.
• $p(\lambda)$ has degree n with n (possibly complex) roots.
• The multiplicity(algebraic) of an eigenvalue is the number of times the associated root of $p(\lambda)$ is repeated.

Similarity

• If A = PBP^-1 for some invertible P, then
  A is similar to B .
• Similar matrices have the same eigenvalues.
• If A = PBP^-1 and $A{\bf x}= \lambda{\bf x}$,
  then $B{\bf y}= \lambda{\bf y}$with ${\bf y}= P^{-1}{\bf x}$.

DIAGONALIZATION

Definition:

A square matrix A is diagonalizable
if A is similar to a diagonal matrix D ( A = PDP^-1).

Theorems:

Let A be an $n\times n$ matrix.

• A is diagonalizable iff A has n linearly independent eigenvectors;
  if A = PDP^-1 for a diagonal matrix D, then the diagonal entries of D are eigenvalues for A with eigenvectors in the respective columns of P.
• If A has n distinct eigenvalues, then A is diagonalizable.
• If A has p distinct eigenvalues, B_k is a basis for the eigenvectors for $\lambda_k$, and $B = B_1 \cup B_2 \dots \cup B_p$,
  then B is a linearly independent set in Rⁿ, and
  A is diagonalizable iff dim(Span(B)) = n.

Diagonalization Method:

For an $n\times n$ matrix A

1.

find the eigenvalues for A, $\lambda_1, \lambda_2, \ldots, \lambda_n$,

2.

find the eigenvectors for A, $\{{\bf v}_1, {\bf v}_2, \ldots, {\bf v}_n\}$, and

3.

construct $P = [{\bf v}_1\ {\bf v}_2\ \ldots\ {\bf v}_n]$ and

$$D = \left[ \begin{array}{cccc} \lambda_1 & 0 & \ldots & 0 \\ ... ...s & \vdots \\ 0 & 0 & \ldots & \lambda_n \end{array}\right] .$$