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연세대 선형대수학 족보 2학기-선대시험-3차기말-모범답안

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작성일18-05-17 11:09

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연세대 선형대수학 족보 2학기-선대시험-3차기말-모범답안 , 연세대 선형대수학 족보 2학기-선대시험-3차기말-모범답안기타시험족보 , 연세대 선형대수학 족보 학기 선대시험 차기말 모범답안

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시험족보/기타




연세대,선형대수학,족보,학기,선대시험,차기말,모범답안,기타,시험족보


Problem 1. Indicate whether the statement is true(T) or (5) If A is a symmetric matrix, then eigenvectors from dierent eigenspaces are orthogonal. (T) false(F). Justify your answer. [each 3pt] (1) If T : Rn → Rn is a linear operator, and if [T ]B = [T ]B with respect to two bases B and B for Rn , then B = B . (F) Solve If T is a zero operator, then [T ]B = O for any basis for R . So [T ]B = [T ]B but B = B . So (λ1 λ2 )(x1 x2 ) = 0 and thus x1 x2 = 0.
n

solve Suppose that x1 ∈ Eλ1 and x...


Problem 1. Indicate whether the statement is true(T) or (5) If A is a symmetric matrix, then eigenvectors from dierent eigenspaces are orthogonal. (T) false(F). Justify your answer. [each 3pt] (1) If T : Rn → Rn is a linear operator, and if [T ]B = [T ]B with respect to two bases B and B for Rn , then B = B . (F) Solve If T is a zero operator, then [T ]B = O for any basis for R . So [T ]B = [T ]B but B = B . So (λ1 λ2 )(x1 x2 ) = 0 and thus x1 x2 = 0.
n

solve Suppose that x1 ∈ Eλ1 and x2 ∈ Eλ2 are eigenvectors from dierent eigenspaces. Then, (λ1 x1 ) x2 = (Ax1 ) x2 = x1 (AT x2 )

= x1 (Ax2 ) = x1 (λ2 x2 )

(2) If V and W are distinct subspaces of Rn with the same dimension, then neither V nor W is a subspace of the other. (T) Solve With out loss of generality, if V is a subspace of W . Since dim(V ) = dim(W ) and V is a subspace of W , V = W . It is a contradiction. Problem 2. Indicate whether the statement is true(T) or false(F). [each 2pt] (1) If A = U ΣV T is a sin…(투비컨티뉴드 ) gular value decomposition of A, then U orthogonally diagonalizes AAT . (T)

(3) If T : Rn → Rn is a one-to-one linear operator, and if {v1 , . . . , vn } is a basis for R , then {T (v1 ), . . . , T (vn )} is also a basis for Rn . (T) Solve We show that {T (v1 ), . . . , T (vn )} is linearly independent. Let c1 T (v1 ) + . . . + cn T (vn ) = 0. Since T is linear, so T (c1 v1 + . . . + cn vn ) = 0 and hence c1 v1 + . . . + cn vn ∈ ker(T ). Since T is 1-1, that is, ker(T ) = {0}, c1 v1 + . . . + cn vn ∈ ker(T ), that is, c1 v1 + . . . + cn vn = 0. Since {v1 , . . . , vn } is linearly independent, c1 = . . . = cn = 0, and hence {T (v1 , . . . , T (vn )} is linearly independent. Therefore {T (v1 , . . . , T (vn )} is a basis for Rn .
n

(2) If S = {v1 , v2 , . . . , vk } is a linearly independent set in Rn , then so is every nonempty subset of S. (T)

(3) Let A, B, C, and D be n × n square matrices. If A is similar to C and B is similar to D, then A + B is similar to C + D. (F)

(4


연세대 선형대수학 족보 2학기-선대시험-3차기말-모범답안
연세대 선형대수학 족보 2학기-선대시험-3차기말-모범답안


Download : 연세대 선형대수학 족보 2학기-선대시험-3차기말-모범답안.pdf( 26 )




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