Singular Value Decomposition Homework: Examining Least Squares Problems in Engineering, Assignments of Electrical and Electronics Engineering

Instructions for a homework assignment on the singular value decomposition (svd) of a matrix as a method for examining least squares problems in engineering. The assignment involves using matlab to compute the svd of a given matrix and using the results to find the solution of a linear system of equations. The document also discusses the significance of the singular values in the context of the matrix and the information they provide.

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Pre 2010

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ECE 497-3: Inverse Problems in Engineering
Homework #2
Due: Friday December 20, 2002
Motivation
In this homework, we will look at the singular value decomposition of a matrix as a method
for examining least squares problems. In this homework, the attatched Matlab routines will
do most of the work. You will need to set the “numeric format” (in the options window) to
“long” for this homework.
1) Consider the following linear system of equations
y=
0.2042 0.0559 0.3715 0.2077
0.1347 0.2208 0.5294 0.1336
0.4803 0.3889 0.6862 0.4670
0.6078 0.6088 1.2159 0.6081
x
Using Matlab, compute yfor xT= [1 1 1 1]. Then, using this y, compute xby determining
the inverse of the above matrix.
It turns out that for any matrix ZRm×nwe can always compute its singular value decom-
position
Z=USVT
where, URm×mand Uis a unitary matrix (UT=U1), VRn×nand Vis a unitary
matrix (VT=V1), and SRm×n.Sis all zeros except for a submatrix, which will will
denote Σ, of size m×mor n×n(depending on whether mor nis smaller) in the upper left
corner. The entries in the submatrix Σ are called the singular values of the matrix and will
b e denoted σifor the ith singlular value. In general, σiσi+1 , that is, the singular values
decrease as the mode number increases (this is important!) As you will see, these singular
values contain a great deal of information about the matrix (and what may be going wrong).
2) Lets assume that
x=Vα
y=Uβ
That is, xcan be written as a linear combination of the columns of V
x=Vα
= [v1v2... vn]α
=v1α1+v2α2+... +vnαn
Since xRnand the columns of Vspan Rn, we know such an expansion must exist. Simi-
larly, we must be able to represent yas a linear combination of the columns of U.
1
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ECE 497-3: Inverse Problems in Engineering Homework #

Due: Friday December 20, 2002

Motivation In this homework, we will look at the singular value decomposition of a matrix as a method for examining least squares problems. In this homework, the attatched Matlab routines will do most of the work. You will need to set the “numeric format” (in the options window) to “long” for this homework.

  1. Consider the following linear system of equations

y =

  

   x

Using Matlab, compute y for xT^ = [1 1 1 1]. Then, using this y, compute x by determining the inverse of the above matrix.

It turns out that for any matrix Z ∈ Rm×n^ we can always compute its singular value decom- position Z = USVT

where, U ∈ Rm×m^ and U is a unitary matrix (UT^ = U−^1 ), V ∈ Rn×n^ and V is a unitary matrix (VT^ = V−^1 ), and S ∈ Rm×n. S is all zeros except for a submatrix, which will will denote Σ, of size m × m or n × n (depending on whether m or n is smaller) in the upper left corner. The entries in the submatrix Σ are called the singular values of the matrix and will b e denoted σi for the ith^ singlular value. In general, σi ≥ σi+1, that is, the singular values decrease as the mode number increases (this is important!) As you will see, these singular values contain a great deal of information about the matrix (and what may be going wrong).

  1. Lets assume that

x = Vα y = Uβ

That is, x can be written as a linear combination of the columns of V

x = Vα = [v 1 v 2 ... vn] α = v 1 α 1 + v 2 α 2 + ... + vnαn

Since x ∈ Rn^ and the columns of V span Rn, we know such an expansion must exist. Simi- larly, we must be able to represent y as a linear combination of the columns of U.

Now assume we have our linear problem y = Zx.

Show that the singular value decomposition leads to

α = Σ−^1 β

(Note, if there are more αi than βi, the extra αi are set to zero. This just means that V spans a larger space than U. Similarly if there are more βi than αi, the extra βi are set to zero. This just means U spans a larger space than V.) This can be written in terms of components as αi = βi/σi

You may assume that S has full rank (.e., S−^1 exists) for this derivation, even if it may not.

  1. Now assume we have an overdetermined least squares solution

xˆ = (ZT^ Z)−^1 ZT^ y

with the assumptions above, and writing Z in terms of its singular value decompostion, show that the solution of the overdetermined least squares problem

xˆ = (ZT^ Z)−^1 ZT^ y

again leads to α = Σ−^1 β

or αi = βi/σi

You may assume that S has full rank (.e., S−^1 exists) for this derivation, even if it may not.

  1. Now we need to get either the α, or, equivalently, the β. The answer is fairly simple. Since we observe y and U is a unitary matrix, you should be able to show that

β = UT^ y

  1. Now compute x using the singular value decomposition of Z. (The Matlab routine will do this for you, it should be the same as in (1).)

  2. Now assume we have a slight error in our measurement, a very likely thing in the “real” world. The Matlab routine will add a small amount of noise to the observation vector. It should usually change the y values by a few percent. Determine the new y when noise is added, and compute a new estimate of x.

  3. Are the estimates in (1) and (6) above close? Do these answers seem strange? Run your program at least three times, with three different sets of noise and turn them