If u + v are eigenvectors with eigenvalue 2 of the matrix A then u – v is an eigenvector of A with eigenvalue 2.

Let M be a 2 × 2 matrix with eigenvalues ₁ = -1.2, A2 = 1 with corresponding eigenvectors Consider the difference equation with initial condition Xo = 5 3 V₁ = V2 = 2 xk+1 = Mxk Write the initial condition as a linear combination of the eigenvectors of M. That is, write x0 = c1V1 + C₂ V₂ = In general, xk= V1+ V2 ) * vi+ k ) v2 Specifically, x4 = For large k, xk≈ k

A diagonalization of the matrix A is given in the form P-¹AP = D. List the eigenvalues of A and bases for the corresponding eigenspaces. (Repeated eigenvalues should be entered repeatedly with the same eigenspaces.) d₁= = ^₂ = 13 = 1007240 18 1007/N +10 18 18 2 1 4 1 2 5 9 4 4 4 0 0 1 4 4 3 9 1 0 1 -1 1 8 0 -1 has eigenspace span has eigenspace span has eigenspace = 80 0 00 0 0 0 -1 (LF) span (smallest λ-value) (largest λ-value)

is an eigenvector of the matrix 2 What is its eigenvalue? Eigenvalue =