Discussion of Exam 2 (see p:\data\math\hartlaub\linearalgebra\exam2.mtw)

Eigenvalues and eigenvectors

Consider an n x n matrix A. A nonzero vector v in R^n is called an eigenvector of A if Av is a scalar multiple of v. That is, Av = (lamda)v, for some scalar lamda. Note that the scalar lamda may be zero. The scalar lamda is called the eigenvalue associated with the eigenvector v.

Important Question

How can we find eigenvalues and eigenvectors of an n x n matrix A?

Note lamda is an eigenvalue of A if and only if there is a nonzero vector v such that Av = (lamda) v.

This equation can be rewritten as Av - (lamda) v = 0 or Av - (lamda) I v = 0.

Notice that this is equivalent to saying that Ker(A - lamda I) is not equal to zero.

This statement is equivalent to saying that the matrix A - lamda I fails to be invertible or the determinant of A - lamda I is equal to zero.

In short, finding an eignevalue is equivalent to solving the characteristic equation det(A - lamda I) = 0.

Discrete Dynamical Systems

Consider the dynamical system x(t+1) = A x(t) with the initial condition x(0) = x_0. Then x(t) = A^t x_0.

If we can find a basis of R^n that consists of the eigenvectors of A, then the state of the system at time t, x(t), can be written as a linear combination of the eigen values, eigenvectors, and coordinates of the initial vector x_0 with respect to the basis formed by the eigenvectors. (See Fact 7.1.3)

Another characterization of an invertible matrix

An n x n matrix is invertible if (and only if) 0 fails to be an eigenvalue of A.

Group/Class Exercises

Section 7.2 - 1, 3, 8, 11, 15, 17, 22, 28, 39

Section 7.1 - 1, 5, 8, 10, 24, 27, 31, 38, 51

Please read Section 7.1-7.4 for class on Thursday