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MA514 - HW 2 - Solved
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 AT (we do not consider A∗ in case this is different from AT):
a11 a21 ... am1
AT = a12 a22 ... am2
... ... ... ... a1n a2n ... amn
We can reverse the ordering of the columns of AT 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 ... amn 1 0 ... 0 amn ... 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 AT = (V ∗)TΣTUT = (V ∗)TΣUT.
This shows that A has the same singular values as AT (even if the left and right singular vectors may change). Also, since P is an orthogonal matrix, we have (ATP)(ATP)∗ = ATPP∗(AT)∗ = AT(AT)∗. This implies that the singular values of ATP are the same as AT (since these are the square roots of the eigenvalues of AT(AT)∗ = (ATP)(ATP)∗. But since ATP = B, we have shown that the singular values of ATP = 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∗ = AAT = UΣΣ∗U∗ = UΣ2UT, we have that UΣUT is a diagonalization of AAT. 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, UTA = ΣV T so that ATUΣ−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:
.
