a) By multiplying A times u show that u is a eigenvector. Whatis ? b) What are t

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a)   A = u vT     => A u   = uvT u = ( vT u ) u    , where    vTu   is a scalar , let =vT u ,    so , we have : A u = u   => = vT u is an eigenvalue of A and it correspondingeigenvector is u . b) Since A has rank one , soit has exactly one nonzero eigenvalue which is vTu    and all other eigenvalues are zero. c) trace (A) = sum of eigenvalues = vT u + 0 = vTu also , A = uvT so the ith diagonal entry of A is uivi    => trace ( A) = uivi =   uT v = vT u

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