Question

With explanation and examples

(a) True or False: If vy is an eigenvector of A with eigenvalue A, then v\ is also an eigenvector of A2 3-13. (b) True or False: If vx is an eigenvector of A with eigenvalue X and A is invertible, then va is also an eigenvector of A-1. (c) It is known that the product of the eigenvalues of a square matrix is the determinant of that matrix. True or False: A matrix with a zero eigenvalue is always invertible (d) True or False: If Av Av\, and B is another n x n matrix satisfying Bva = uv^, then v\ is an eigenvalue for A+ B

Answer 1

3-13. (a) TRUE. If v is an eigenvector of A associated with the eigenvalue ʎ, then Av = ʎv so that A² v =A(Av) = A( ʎv) = ʎ (Av) = ʎ² v. This means that v is an eigenvector of A2 associated with the eigenvalue ʎ².

(b). TRUE. If v is an eigenvector of A associated with the eigenvalue ʎ, then Av = ʎv . On multiplying to the left by A^-1, we get A^-1Av = A^-1(ʎv) or, ʎ A^-1v   = v or, A^-1v   == (1/ ʎ)v. Thus, v is an eigenvector of A^-1 associated with the eigenvalue 1/ʎ.

( c). FALSE. If 0 is an eigenvalue of A, then det(A) = 0 so that A is not invertible.

(d). TRUE. If v is an eigenvector of A associated with the eigenvalue ʎ and if v is an eigenvector of B associated with the eigenvalue μ, then Av = ʎv and Bv = μv so that (A+B)v = Av+bv = ʎv+ μv = (ʎ+μ)v. This means that v is an eigenvector of A+B associated with the eigenvalue ʎ+μ.

Note: There is a mispintin part(d). v is an eigenvector of A+B and not an eigenvalue.