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1. Given the SVD of a matrix A ∈ Cm×n
A = UΣVH (1)
with the unitary matricesr×r U ∈ Cm×m, V ∈ Cn×n and the positive definite diagonal matrix Σ1 ∈ R , where r = rank(A), the Moore-Penrose pseudo inverse A+ of A is defined as
A (2)
(a) What are the dimensions of the matrices U1, U2, V1, V2, Σ and A+ ?
(b) Show that U and V
(c) 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)
2. Look at the matrix A and its pseudo inverse A+
A ; A
in terms of four real numbers a,b,c,d ∈ R.
(a) Write down the three linear equations of the variables a,b,c,d that follow from (3), (5) and (6) in the form:
D e (7)
(Hint: since we are dealing with real numbers, the (•)H operator becomes a pure transposition)
(b) Compute the general solution of the underdetermined system (7) in terms of d.
(c) Now determine the unique value d, that will also satisify the last Moore-Penrose condition (4)
(d) Write down the pseudo inverse A+ you have obtained in this way.
2
(e) A SVD of A is given as
Compute A+ by the definition in (2) and compare to the previous result.
3. Consider the matrix
P (8)
(a) Show that P is a projector onto a vector space.
(b) This vector space S = range(P) is a subspace of C3. What is its dimension?
(c) What is the dimension of the orthogonal complement S⊥ ⊂ C3 of S in C3?
(d) Compute the projector P⊥ onto S⊥
(e) Using this result compute an orthonormal base of S⊥
(f) Have a look at the following subspace
S2 = range (9)
Is S2 = S ? (Hint: Compute the projector onto S2 and compare to (8).)
4. The system Ax = b with A ∈ Cm×n, 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
(a) the least-squares solution obtained by the pseudo inverse, i.e. xLS = A+b, is really an exact solution to the modified system Ax Pb
(b) the error of the righthand sides is orthogonal to bb.