线性代数的题 第三问和第四问该如何证明
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3) Since v4 nonzero and A^T*v4 = 0, A*A^T*v4 = 0 = 0*v4 which means that v4 is an eigenvector corresponding to eigenvalue 0.
Note that (A^T*A)*ui = di*ui, where ui is the i-th column of U and di is the i-th diagonal element of D. Therefore A*A^T*vi = A*(A^T*A*ui) = A*di*ui = di*vi that's to say, vi is an engenvector of A*A^T corresponding to eigenvalue di.
4) I suspect that eigenvalue of A^T*A are different. (From (1))
Eigenvectors corresponding to different eigenvalues are linearly independent. Therefore (v1 ... v4) are linearly independent and is a basis of R^4.
Note that (A^T*A)*ui = di*ui, where ui is the i-th column of U and di is the i-th diagonal element of D. Therefore A*A^T*vi = A*(A^T*A*ui) = A*di*ui = di*vi that's to say, vi is an engenvector of A*A^T corresponding to eigenvalue di.
4) I suspect that eigenvalue of A^T*A are different. (From (1))
Eigenvectors corresponding to different eigenvalues are linearly independent. Therefore (v1 ... v4) are linearly independent and is a basis of R^4.
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