If T maps Vn into Vn, then x and y = T(x) are vectors in the same n−dimensional vector spa...

If T maps Vn into Vn, then x and y = T(x) are vectors in the same n−dimensional vector space and can be compared (see Problems 9, 10, and 11). If T(x) is a scalar multiple of x, for some nonzero x, say T(x) = ax, then we say that x is an eigenvector of T, associated with the eigenvalue λ. If T has the matrix A, then x is also an eigenvector of A, associated with the eigenvalue λ, as in Section 1.11.

Prove: If v1, …, vk are eigenvectors of the linear mapping T,associated with distinct eigenvalues λ1, ..., λ2, then v1, ..., vk are linearly independent. [Hint: Let T have the matrix A. Use induction. Verify that the assertion is true for k: = 1. Assume that it is true for k = m < n and prove it for k = m + 1 by assuming multiplying by A on the left, and then eliminating vm + 1 to obtain

Now use the induction hypothesis to conclude that c1 = 0,..., cm = 0 and hence also Cm +1 0.]