x5.1 Eigenvectors and Eigenvalues The goal in this chapter is to dissect the action of a linear transformation x 7! Ax into elements that are easily visualized. (NB: For most of Chapter 5, A is n × n.) What we seek: vectors that stretch (dilate) or shrink (contract), but don't rotate, under the linear n transformation x 7! Ax. In other words, which vectors x 2 R satisfy Ax is parallel x? dfn: An eigenvector of an n × n matrix A is a nonzero vector x such that Ax = λx for some scalar λ. The scalar λ is called the eigenvalue associated with eigenvector x. Comments: • Note the geometry: if x is an eigenvector of A, then Ax dilates or contracts x, but does not rotate it. • If the eigenvector x has corresponding eigenvalue λ, then any non-zero v 2 Spanfxg is an eigenvector with that same eigenvalue λ. • The eigenvector x can never be 0, but the eigenvalue λ can be 0. • Only square matrices have eigenvectors and eigenvalues. Finding eigenvectors: Given an eigenvalue λ of A, we need to find all nonzero x such that Ax = λx. This is equivalent to solving the homogeneous equation Ax − λx = 0 which we can turn into a matrix difference by using the fact that if In is the n × n identity matrix, then Inx = x. Thus Ax − λx = 0 becomes Ax − λInx = 0 which simplifies to (A − λIn)x = 0. The nonzero solutions of (A − λIn)x = 0 are the eigenvectors corresponding to the given eigenvalue λ. The full set of solutions to this equation (i.e. including 0) comprise the null space of the n × n matrix n (A − λIn). This null space is a subspace of R called the eigenspace of A corresponding to λ. Briefly: the eigenspace of A corresponding to λ is Nul(A − λIn). It consists of the eigenvectors of A corresponding to λ together with 0. Moral: To find eigenvalues of A, solve the characteristic equation det(A − λIn) = 0 for λ. If λ is an eigenvalue, then to find the eigenspace solve (A − λIn)x = 0; that is: find the null space of (A − λIn). The eigenvectors associated to λ are the nonzero vectors in this eigenspace. n NB: If λ is an eigenvalue of A then the eigenspace of A associated to λ is a subspace of R since it is Nul(A − λIn). NB: If a particular matrix A has eigenvalue λ = 0, there are nonzero eigenvectors x with Ax = 0 · x = 0: n What subspace of R do these vectors form? Nul A !!! So if A has eigenvalue λ = 0, Nul A 6= 0 hence A is n n not invertible and the map x 7! Ax from R to R cannot be one-to-one nor onto. Theorem 1 The eigenvalues of a triangular matrix are the entries on its main diagonal. Theorem 2 If v1; v2;:::; vr are eigenvectors that correspond to distinct eigenvalues λ1; λ2; : : : ; λr of an n × n matrix A, then the set fv1; v2;:::; vrg is linearly independent. n Some observations: When A is n × n the map x 7! Ax is a map from R to itself. If x is in the eigenspace of A corresponding to λ, then multiplication by A keeps that vector x in the eigenspace because Ax = λx and vector subspaces, in particular eigenspaces, are closed under scalar multiplication. We say that the eigenspace corresponding to λ is invariant under multiplication by A because an eigenvector in that eigenspace is always mapped to an eigenvector in that same eigenspace (if x is an eigenvector associated to λ, then Ax = λx)..