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 $\left[\begin{array}{c}{1 / \sqrt{6}} \\ {-2 / \sqrt{6}} \\ {1 / \sqrt{6}}\end{array}\right].$] $\left[\begin{array}{ll}{7} & {1} \\ {5} & {5} \\ {0} & {0}\end{array}\right]$

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 $\left[\begin{array}{c}{1 / \sqrt{6}} \\ {-2 / \sqrt{6}} \\ {1 / \sqrt{6}}\end{array}\right].$]
$\left[\begin{array}{ll}{7} & {1} \\ {5} & {5} \\ {0} & {0}\end{array}\right]$

Key Concepts

Singular Value Decomposition (SVD)

The SVD is a factorization of a real or complex matrix into three matrices, A = U?V?, where U and V are orthogonal (or unitary) matrices and ? is a diagonal matrix containing the singular values. This decomposition provides insight into the geometric structure of the matrix, including its range, null space, and rank, and is useful for solving systems of equations, optimization, and data analysis.

Orthogonal (Unitary) Matrices

An orthogonal (or unitary in the complex case) matrix has the property that its columns (and rows) form an orthonormal basis, meaning each column vector has unit length and is mutually perpendicular to the others. In the context of the SVD, the matrices U and V are orthogonal, which ensures that the decomposition preserves lengths and angles, making it a powerful tool in numerical linear algebra.

Singular Values

The singular values in the SVD are the non-negative entries along the diagonal of the ? matrix. They represent the ‘strength’ or ‘importance’ of the corresponding singular vectors in the matrices U and V and are typically arranged in descending order. These values are found by taking the square roots of the eigenvalues of A?A (or AA?) and provide critical information about the inherent dimensionality and conditioning of the matrix.

Eigenvalue Decomposition Connection

The process of finding the SVD involves computing the eigenvalue decompositions of A?A and AA?. The eigenvectors of A?A form the right singular vectors (columns of V) and the eigenvectors of AA? form the left singular vectors (columns of U). The singular values are the square roots of the eigenvalues from these decompositions, linking the SVD to classical eigenvalue problems in linear algebra.

Transcript

00:01 Okay, so for this problem, we first consider the matrix a transpose times a.

00:09 So in this case, this new matrix equals to 74, 32, 32, and 26.

00:18 So the step one for the singular value decomposition for matrix a will be finding out all of the eigenvalues and eigenvectors of this matrix.

00:28 The characteristic equation for this metric is the following.

00:33 And the determinant of a transpose times a minus lambda i equals to 0, which gives us a quadratic equation, 74 minus lambda times 26 minus lambda minus 32 square equals to 0.

00:49 By solving this quadratic equation, we have two solutions, lambda 1 equals to 90, and lambda 2 equals to 10.

00:56 Okay, so with this result, we can directly write out the singular values of a.

01:05 Sigma 1 equals to root of 90.

01:08 Sigma 2 equals to root of 10.

01:10 So in the singular value decomposition of matrix a, the matrix sigma equals to a diagonal matrix with the same size as the original matrix a.

01:25 So it looks like this.

01:28 Now we are looking for the eigenvectors of this matrix.

01:33 A transpose times a minus lambda 1i equals to the following matrix.

01:40 And we can directly write it out, write out the eigenvectors of this matrix.

01:50 Now we'll be 2 over root of 5 and 1 over root of 5...

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