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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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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.
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).
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.
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.
β = UT^ y
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).)
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.
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