[SOLVED] MA514 -HW 2

Description

Exercise 4.1

(a) Determine the SVD of the matrix:

 .

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

AA^∗ = UΣV ^∗V Σ^∗U^∗ = UΣΣ^∗U^∗ A^∗A = V Σ^∗U^∗UΣV ^∗ = V Σ^∗ΣV ^∗ .

In particular, UΣΣ^∗U^∗ and V Σ^∗ΣV ^∗ are a diagonalizations of A^∗A (note that ΣΣ^∗ and Σ^∗Σ are diagonal matrices. Therefore we can determine Σ and V by determining the eigenvalues of A^∗A and finding corresponding eigenvectors.

Since A^∗A is diagonal, the eigenvalues are the diagonal elements. The eigenvalues of A^∗A 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 A^∗A is diagonal, we can find eigenvectors by inspection as v₁ = e₁ and v₂ = e₂. So we have

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

√

Since Σ scales the first column of U by           σ₁ = 3 and scales the second

√

column by       σ₂ = 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 …            am10        …      0             1 am1 …              a21 a11

ATP = _a¹² a²² …                 am2__0    …      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 = UΣV ^∗, we see that A^T = (V ^∗)^TΣ^TU^T = (V ^∗)^TΣU^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 = UΣV ^T such that one has the minimal number of minus signs in U and V . Using AA^∗ = AA^T = UΣΣ^∗U^∗ = UΣ²U^T, we have that UΣU^T is a diagonalization of AA^T. So we will find the eigenvalues,  and of Σ² and corresponding normalized eigenvectors to form U.

Next find an eigenvector for

!

Next find an eigenvector for

!

We takeand Σ =.

Then since A = UΣV ^T, U^TA = ΣV ^T so that A^TUΣ⁻¹ = V . We can use this to find V :

!

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

 .

√                         √

(b) The singular values of A are σ₁ = 10 2 and σ₂ = 5 2.

The left singular vectors of A are:

 .

The right singular vectors of A are:

 .