Question: Let x. y be nonzero sectors in Rn. n 2. and let A = xyT Show that (a) A = 0 is an eigenvalue

Let x. y be nonzero sectors in Rn. n ≥ 2. and let A = xyT Show that
(a) A = 0 is an eigenvalue of A with n - I linearly independent elgenvectors and consequently has multiplicity at least n - 1 (see Exercise 15).
(b) The remaining elgenvalue of A is
λn = tr A= xTy
and x is an eigenvector belonging to λn.
(c) if λn = xTY ≠ then A is diagonalizable.

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