Let x, y be nonzero vectors in Rn, n 2, and let A = xyT . Show that (a) = 0 is an

Chapter 6, Problem 17

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Let x, y be nonzero vectors in Rn, n 2, and let A = xyT . Show that (a) = 0 is an eigenvalue of A with n 1 linearly independent eigenvectors and consequently has multiplicity at least n1 (see Exercise 16). (b) the remaining eigenvalue of A is n = tr A = xT y and x is an eigenvector belonging to n. (c) if n = xT y _= 0, then A is diagonalizable. 1

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