(a) Determine the SVD of the matrix:

.

We want a decomposition of the form A = UV , where U and V are unitary matrices and a diagonal matrix containing the square roots of the singular values of A. We then know that

AA = UV V U = UU AA = V UUV = V V .

In particular, UU and V V are a diagonalizations of AA (note that and are diagonal matrices. Therefore we can determine and V by determining the eigenvalues of AA and finding corresponding eigenvectors.

Since AA is diagonal, the eigenvalues are the diagonal elements. The eigenvalues of AA are the squares of the singular values of A. Thus,

= 4 and next we find corresponding normalized eigenvectors to use a columns of V (so that V will be unitary). In this case, since AA is diagonal, we can find eigenvectors by inspection as v₁ = e₁ and v₂ = e₂. So we have

Next, use the requirement that AV = U to find U.

Since scales the first column of U by ₁ = 3 and scales the second

column by ₂ = 2, so we have

.

We conclude that an SVD factorization of A is given by

Exercise 4.2

Suppose A is an m n matrix and B is the n m matrix obtained by rotating A ninety degree clockwise on paper. We prove that A and B have the same singular values.

Let A be written as

a11 a12 a1n

A = a²¹ a²² a2n

am1 am2 amn

Consider A^T (we do not consider A in case this is different from A^T):

a11 a21 am1

AT = a¹² a²² am2

a1n a2n amn

We can reverse the ordering of the columns of A^T by multiplying by the m m matrix P that has 1s on the antidiagonal and 0s elsewhere:

a11 a21 am10 0 1 am1 a21 a11

ATP = a¹² a²² am20 1 0 = am² a²² a12 = B

a₁n a₂n a_mn 1 0 0 a_mn a₂n a₁n Therefore, we have found the matrix equation that relates the matrix A and B from the given description of how B is obtained from A. Next, since A has an SVD A = UV , we see that A^T = (V )^TTU^T = (V )^TU^T.

This shows that A has the same singular values as A^T (even if the left and right singular vectors may change). Also, since P is an orthogonal matrix, we have (A^TP)(A^TP) = A^TPP(A^T) = A^T(A^T). This implies that the singular values of A^TP are the same as A^T (since these are the square roots of the eigenvalues of A^T(A^T) = (A^TP)(A^TP). But since A^TP = B, we have shown that the singular values of A^TP = B are the same as A. Thus, the singular values of A are the same as the singular values of B, as desired.

Exercise 4.3

See the MATLAB script for this exercise.

Exercise 5.3

Consider the matrix

.

(a) We will determine a real SVD of A of the form A = UV ^T such that one has the minimal number of minus signs in U and V . Using AA = AA^T = UU = U²U^T, we have that UU^T is a diagonalization of AA^T. So we will find the eigenvalues, and of ² and corresponding normalized eigenvectors to form U.

Next find an eigenvector for

!

Next find an eigenvector for

!

We takeand =.

Then since A = UV ^T, U^TA = V ^T so that A^TU¹ = V . We can use this to find V :

!

Therefore we have a factorization A = UV ^T with:

.

(b) The singular values of A are ₁ = 10 2 and ₂ = 5 2.

The left singular vectors of A are:

.

The right singular vectors of A are:

.