- Given the SVD of a matrix A Cmn
A = UVH (1)
with the unitary matricesrr U Cmm, V Cnn and the positive definite diagonal matrix 1 R , where r = rank(A), the Moore-Penrose pseudo inverse A+ of A is defined as
A (2)
- What are the dimensions of the matrices U1, U2, V1, V2, and A+ ?
- Show that U and V
- Show that (2) satisfies the four Moore-Penrose conditions for a pseudo inverse
AA+A | = | A | (3) |
A+AA+ | = | A+ | (4) |
AA+ = AA (5)
A+A = A (6)
- Look at the matrix A and its pseudo inverse A+
A ; A
in terms of four real numbers a,b,c,d R.
- Write down the three linear equations of the variables a,b,c,d that follow from (3), (5) and (6) in the form:
De (7)
(Hint: since we are dealing with real numbers, the ()H operator becomes a pure transposition)
- Compute the general solution of the underdetermined system (7) in terms of d.
- Now determine the unique value d, that will also satisify the last Moore-Penrose condition (4)
- Write down the pseudo inverse A+ you have obtained in this way.
2
- A SVD of A is given as
Compute A+ by the definition in (2) and compare to the previous result.
- Consider the matrix
P (8)
- Show that P is a projector onto a vector space.
- This vector space S = range(P) is a subspace of C3. What is its dimension?
- What is the dimension of the orthogonal complement S C3 of S in C3?
- Compute the projector P onto S
- Using this result compute an orthonormal base of S
- Have a look at the following subspace
S2 = range (9)
Is S2 = S ? (Hint: Compute the projector onto S2 and compare to (8).)
- The system Ax = b with A Cmn, x Cn and b Cm has an exact solution only if b range(A). If m > n (i.e. more equations than unknowns) there is usually no exact solution, since b / range(A) in most cases. The best we can do is modify the righthand have a SVD as in (1) and define a projector Pbas side of the system such that b is replaced by b, its projection onto the range of A. Let A
P (10)
Show that
- the least-squares solution obtained by the pseudo inverse, i.e. xLS = A+b, is really an exact solution to the modified system AxPb
- the error of the righthand sides is orthogonal to bb.
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