With explanations please. If you don\'t know one that\'s fine. I\'d rather you s
ID: 3117182 • Letter: W
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With explanations please. If you don't know one that's fine. I'd rather you say you dont know then give me the wrong answer. I'll give a thumbs up
1. (22x1 points) Let A, B,C be square matrices. Answer true (T) or false (F) (a) (T/F) Let u be the zero vector. Then Au 411, so 4 is an eigenvalue of A. (b) (T/F) If the rows of A are linearly independent then det(A)0 (c) (T/F) Not every nonzero subspace of R has an orthonormal basis. (d) (T/F) For every orthonormal matrix B, Bu has the same length as u for all vectors u (e) (T/F) An mxnmatrix D has rank W and any independent vectors v1,..., Vm E V, the images g(vi),g(vm) are independent vectors in WExplanation / Answer
1 (a). False. The equation Au = u, whatever be the value of , does not determine eigenvalues if u = 0.
(b). True. If the rows of A are linearly independent, then A can be row reduced to the identity matrix s that A is invertible and hence det(A) 0
(c ). False. Every non-zero subspace W of Rn ,of dimension m has an orthonormal basis {e1,e2,…,em}.
(d). True. Orthonormal matrices preserve length ( and angles).
(e). True. If a mxn matrix A has rank < m, then the RREF of A has some zero row(s), so that the rows of A are linearly dependent.
(f).True, asdim H + dim H =n.
(g). False. det(AB) = det(A) det(B) so that det ((AB)35) = det(AB).det(AB)…35 times = (det(A)det(B))35 = (det(A))35( det(B))35.
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