MATH125 Lecture Notes - Lecture 33: Eigenvalues And Eigenvectors, Diagonalizable Matrix

• School: University of Alberta
• Department: Mathematics
• Course: MATH125
• Professor: Nikita

Document Summary

The following theorem shows that the eigenvectors p1, p2, p3 from the previous example are automatically linearly independent so that we did not need to check it: , k be distinct eigenvalues of a square matrix a. If bi be a basis of the eigenspace e i, then the set b := b1 bk is linearly independent. , vk of a square matrix a corresponding to distinct eigenvalues are linearly independent. We use induction on k. for k = 1 we have only one eigenvector vector v1. Since v1 = 0, it constitutes a linearly independent set. Now assume that we already proved the statement for any k vectors and let us prove it for k + 1 vectors v1, . For this, assume that c1v1 + + ckvk + ck+1vk+1 = 0 for some scalars c1, . , ck, ck+1 and prove that then c1 = = ck = ck+1 = 0.