Find an SVD of each matrix [Hint: In Exercise 11, one choice...

Find an SVD of each matrix [Hint: In Exercise 11, one choice for $U$ is $\left[\begin{array}{rrr}{-1 / 3} & {2 / 3} & {2 / 3} \\ {2 / 3} & {-1 / 3} & {2 / 3} \\ {2 / 3} & {2 / 3} & {-1 / 3}\end{array}\right]$ In Exercise $12,$ one column of $U$ can be $\left[\begin{array}{c}{1 / \sqrt{6}} \\ {-2 / \sqrt{6}} \\ {1 / \sqrt{6}}\end{array}\right].$] $\left[\begin{array}{rr}{3} & {-3} \\ {0} & {0} \\ {1} & {1}\end{array}\right]$

Find an SVD of each matrix [Hint: In Exercise 11, one choice for $U$ is $\left[\begin{array}{rrr}{-1 / 3} & {2 / 3} & {2 / 3} \\ {2 / 3} & {-1 / 3} & {2 / 3} \\ {2 / 3} & {2 / 3} & {-1 / 3}\end{array}\right]$ In Exercise $12,$ one column of $U$ can be $\left[\begin{array}{c}{1 / \sqrt{6}} \\ {-2 / \sqrt{6}} \\ {1 / \sqrt{6}}\end{array}\right].$]
$\left[\begin{array}{rr}{3} & {-3} \\ {0} & {0} \\ {1} & {1}\end{array}\right]$

Key Concepts

Relationship Between SVD and Eigenvalue Problems

The computation of singular values is intimately connected with eigenvalue problems since the singular values are the square roots of the eigenvalues of A^T A (when A is a real matrix). This relationship allows techniques from spectral theory to be applied to the analysis of the SVD, tying together concepts of eigenvectors, eigenvalues, and orthogonal diagonalization in the process of decomposing the matrix.

Singular Value Decomposition (SVD)

Singular value decomposition is a factorization of a matrix into a product of three matrices: U, ?, and V^T. The matrix U contains orthonormal columns and accounts for the domain, ? is a diagonal matrix with nonnegative entries (the singular values), and V^T contains orthonormal rows. This decomposition provides insight into the geometric properties of the matrix, such as its range and null space, and is applicable to any m×n matrix.

Orthogonal Matrices

Orthogonal matrices are square matrices whose columns (and rows) form an orthonormal set. In the context of SVD, both U and V are orthogonal matrices, meaning that U^T U and V^T V yield the identity matrix. This property is essential because it ensures that the transformation preserves lengths and angles, and it simplifies many computations involving the matrix decomposition.

Singular Values

Singular values are the nonnegative entries on the diagonal of the ? matrix in the SVD. They are determined as the square roots of the eigenvalues of A^T A (or AA^T) and provide a measure of the magnitude of the action of the matrix in different directions. These values are crucial in understanding the behavior of the matrix, including its rank, condition number, and stability properties.

Transcript

00:01 Okay, so in this problem we first write out metrics we are looking for.

00:07 So the matrix should be a transpose times a.

00:11 So in this case, this matrix equals to 10 minus 8, minus 8, 10.

00:20 And the step one is finding out the eigenvalues of this matrix.

00:28 So the characteristic equation for this matrix will be a transpose times a minus lambda i.

00:33 Equals 0, which is 10 minus lambda square minus 64 equals 0.

00:41 This quadratic equation has two solutions, lambda 1 equals to 18, and the lambda 2 equals to 2.

00:48 So the singular values for matrix a will be sigma 1 equals to 3 times root of 2, and the sigma 2 equals to root of 2.

00:58 That means in the singular value decomposition of matrix a, the matrix sigma will be 3 root of 2, 0, 0 root of 2, but this is not enough.

01:12 This matrix sigma sure has, have the same size as the matrix a.

01:17 So it should be something like this.

01:19 We use 0s to add, we add an extra row to this matrix, and it will be a 3x2 matrix, which is the same size as the original matrix a.

01:36 And once we figure out these eigenvalues, we can find out the eigenvectors.

01:43 So a transpose times a minus sigma 1i equals to minus 8, minus 8, minus 8, minus 8, minus 8, minus 8, minus 8, minus 8, minus 8, and then we can choose v1 to be 1 over root of 2 and minus 1 over root of 2...

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