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Midterm Solution
Problem 1 1. True. Let Ax = λx with λ ∈ C and non-zero x ∈ Cn. Because A2 = A, Ax = A2x = Aλx = λ2x = λx, which implies that (λ2 − λ)x = 0. Since x contains at least one non-zero element, λ2 − λ = 0 must hold. It follows that λ = 0 or 1. 2. True. Note that the rank of a zero matrix is 0 and its dimension of kernel (null) space is n by the rank-nullity theorem. Besides, Ker(A) = Ker(ATA) (Since if Ax = 0, ATAx = 0. Conversely, if ATAx = 0, Ax = 0 follows from xTATAx = 0). Therefore, by the rank-nullity theorem, rankA = n − dimKer(A) = n − dimKer(ATA) = 0, which implies that A is zero matrix. (An alternative proof is that tr(ATA) = 0 implies m∑ i=1 n∑ j=1 a2ij = 0. Thus aij = 0 where aij denotes the element in the ith row and jth column of A) 3. True. Since A−B is real symmetric, there exist an orthogonal matrixO and a real diagonal matrix D with diagonal elements {d1, d2, ..., dn} such that A − B = OTDO. From xT (A − B)x = 0, xTOTDOx = 0. We can construct a series of vectors xi such that Oxi = ei where ei is the standard unit vector (only i th element is 1 and other elements are 0 in this vector). Since di = e T i Dei = 0 for i = 1, 2, ..., n, we conclude that D = 0 and it follows that A − B = 0 and A = B. Problem 2 1. Let A have distinct eigenvalues λ1, ..., λn with corresponding eigenvectors x1, ..., xn. If follows that Axi = λixi for i = 1, ...n. Since A commutes with B, ABxi = BAxi = Bλixi = λiBxi which implies that Bxi is also an eigenvector of A with respect to eigenvalue λi (when Bxi = 0, the following analysis still holds). Therefore, Bxi lies in the eigenspace spanned by xi so that there exists µi such that Bxi = µixi. Hence (µi, xi) is an eigenvalue and eigenvector pair for B, for each i = 1, ...n. Let O = [x1, x2, ..., xn] and D be a diagonal matrix with diagonal elements {µ1, µ2, ..., µn}, then BO = OD and B = ODO−1. Therefore, B is diagonalizable. 2. Let Λ be a diagonal matrix with diagonal elements λ1, ..., λn, then A = OΛO −1. Since B = ODO−1, it is sufficient to prove that there exist coefficients a0, a1, ..., an−1 such that D = an−1Λn−1+ an−2Λn−2+ ...+ a1Λ+ a0I. Since D and Λ are diagonal matrices, µi = an−1λn−1i + an−2λn−2i + ...+ a1λi+ a0 should hold for i = 1, ..., n. Equivalently, there should exist a solution for the linear equation 1 λ1 · · · λn−11 1 λ2 · · · λn−12 ... ... . . . ... 1 λn · · · λn−1n a0 a1 ... an−1 = µ1 µ2 ... µn 1 where the leftmost square matrix denoted as V is a Vandermonde matrix and det(V ) = ∏ 1≤i(λj − λi) is non-zero since A has distinct eigenvalues. Thus V is non-singular and there exists a unique solution for the linear equation, which completes the proof. Problem 3 1. Characteristic polynomial of A: A− λI = −λ3 + 2λ2 − λ = −λ(λ− 1)2. Therefore, eigenvalues of A: 0 or 1. 2. Algebraic multiplicity for λ = 0 is 1, and for λ = 1 is 2. 3. Geometric multiplicity for λ = 0 is 1, and for λ = 1 is 1. 4. A matrix is diagonalizable if and only if the algebraic multiplicity equals the geometric multi- plicity of each eigenvalue. Therefore, matrix A is not diagonalizable 5. For Av1 = 0, v1 = 0−1 2 , For (A− I)v2 = 0, v2 = 1−1 5 , For (A− I)v3 = v2, v3 = 03 −5 6. P = 0 1 0−1 −1 3 2 5 −5 and rank of matrix P is 3. J = P−1AP = 0 0 00 1 1 0 0 1