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1. Solution: scores (a). True The eigenspace of A corresponding to eigenvalue λ is the null space of A I λ . Since 2 2 A × ∈ℜ and with an eigenspace of dimension 2, rank ( ) 0 A I λ = , or, equivalently, ( ) 0 A I λ = A I λ = . 5 (a). False D is a diagonal matrix and its diagonal entry ii d is the eigenvalue of A corresponding to column i p , eigenvector corresponding to ii d . If we change the permutation of the columns of P , the order of the diagonal entries of D will change. Ex. 1 1 1 3 2 3 1 0 2 3 3 2 2 0 3 2 0 2 0 1 0 2 0 1 1 0 0 1 1 0 A ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ = = = ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 5 2. Solution: scores Assume that n x ^ satisfies 0 Ax = , then ( ) 0 T T A Ax A Ax = = . Thus ( ) ( ) Null Null T A A A (1) Assume that n x ^ satisfies 0 T A Ax = , then ( ) 2 0 T T T T x A Ax x A Ax Ax = = = 0 Ax = . Thus ( ) ( ) Null Null T A A A (2) From (1) and (2), we have ( ) ( ) Null Null T A A A = Nullity of Nullity of T A A A = ( ) ( ) ( ) rank rank Nullity of T A A A n A = = (3) 10

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