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Chapter 6: Problem 1

Show that \(A\) and \(A^{T}\) have the same nonzero singular values. How are their singular value decompositions related?

Short Answer

Expert verified
The eigenvalues of \(A\) and \(A^T\) can be found using the characteristic equation: \(det(A-\lambda I) = 0\) and \(det(A^T-\lambda I) = 0\), and the singular values can be determined by taking the square root of the absolute values of eigenvalues. By comparing the singular values, we can verify that \(A\) and \(A^T\) have the same nonzero singular values. The singular value decompositions of \(A\) and \(A^T\) are related as: \(A = U \Sigma V^T\) \(A^T = V \Sigma^T U^T\) where the matrices \(U\) and \(V\) can be transposed to switch their roles, and the diagonal matrix \(\Sigma\) stays the same, containing the singular values.

Step by step solution

01

Find the eigenvalues of A and A^T

To find the singular values of A and A^T, we first need to find their eigenvalues. The eigenvalues of A and A^T can be found using the characteristic equation: \(det(A-\lambda I) = 0\) \(det(A^T-\lambda I) = 0\) where \(A\) is the matrix, and \(\lambda I\) represents the eigenvalue multiplied by the identity matrix. Solve for \(\lambda\) in both equations.
02

Determine the singular values from the eigenvalues

Once the eigenvalues are found, they can be used to determine the singular values (\(\sigma_i\)) of A and A^T by taking the square root of the absolute value of the eigenvalues of A and A^T, respectively: \(\sigma_i = \sqrt{|\lambda_i|}\) Sort the singular values in decreasing order.
03

Compare the nonzero singular values of A and A^T

Now that we have the singular values of both A and A^T, compare them and verify that they are the same for all nonzero singular values.
04

Analyze the relationship between the singular value decompositions of A and A^T

To analyze the relationship between the singular value decompositions (SVD) of A and A^T, consider their SVD forms: \(A = U \Sigma V^T\) \(A^T = V \Sigma^T U^T\) The matrix \(\Sigma\) has the same singular values for A and A^T, as shown in the previous steps. The columns in the matrices U and V are eigenvectors of AA^T and A^TA respectively. Due to the relationship between A and A^T, they share the same singular values, and their singular value decompositions are related by: \(A^T = V \Sigma^T U^T\) where the matrices U and V can be transposed to switch their roles, and the diagonal matrix \(\Sigma\) stays the same.

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