Defn: A set V together with two operations, called addition Linear Algebra Review: Eigenvalues and Eigenvectors and scalar multiplication is a vector space if the following vector space axioms are satisfied for all vectors u, v, and w Defn: λ is an eigenvalue of the linear transformation in V and all scalars, c, d in R. T : V → V if there exists a nonzero vector x in V such that T (x) = λx. The vector x is said to be an eigenvector Vector space axioms: corresponding to the eigenvalue λ. a.) u + v is in V 4 1 Example: Let T (x) = x.  5 0  b.) cu is in V u v v u c.) + = + 4 1 −1 1 −1 Note = = −1 d.) (u + v) + w = u +(v + w)  5 0  5   −5   5  e.) There is a vector, denoted by 0, in V such that −1 Thus -1 is an eigenvalue of A and is a corresponding u + 0 = u for all u in V  5  eigenvector of A. f.) For each u in V , there is an element, denoted by −u, in V such that u +(−u) = 0 4 1 1 5 1 g.) (cd)u = c(du) Note = = 5  5 0  1   5   1  h.) (c + d)u = cu + du 1 Thus 5 is an eigenvalue of A and is a corresponding i.) c(u + v) = cu + cv  1  eigenvector of A. j.) 1u = u

Examples: 4 1 2 16 2 Note = =6 k for any k. 5 0 8 10 8 1.) Rk with the usual operations of addition and scalar mul-        tiplication is a vector space. 2 Thus is NOT an eigenvector of A. 2.) The set M k,n, the set of all k × n matrices with the usual  8  operations of addition and scalar multiplication is a vector MOTIVATION: Thus to find the eigenvalues of A and their corresponding eigenvectors: 2 −1 1 Note = + 3  8   5   1  Step 1: Find eigenvalues: Solve the equation 2 −1 1 −1 1 det(λI − A) = 0 for λ. Thus A = A( + 3 ) = A + 3A  8   5   1   5   1  Step 2: For each eigenvalue λ0, find its corresponding eigen- −1 1 16 vectors by solving the homogeneous system of equations = −1 + 3 · 5 =  5   1   10  (λ0I − A)x = 0 for x.

Defn: det(λI − A)=0is the characteristic equation of A. Thm 3: The eigenvalues of an upper triangular or lower tri- Finding eigenvalues: angular matrix (including diagonal matrices) are identical to Suppose Ax = λx (Note A is a SQUARE matrix). its diagonal entries. Then Ax = λIx where I is the identity matrix. Defn: The eigenspace corresponding to an eigenvalue λ0 of a matrix A is the set of all solutions of (λ0I − A)x = 0. Thus λIx − Ax =(λI − A)x = 0 Thus if Ax = λx for a nonzero x, then (λI − A)x = 0 has a nonzero solution. Note: An eigenspace is a vector space Thus det(λI − A)x = 0. The vector 0 is always in the eigenspace. Note that the eigenvectors corresponding to λ are the nonzero The vector 0 is never an eigenvector. solutions of (λI − A)x = 0. The number 0 can be an eigenvalue. Thm: A square matrix is invertible if and only if λ = 0 is not an eigenvalue of A.