1. Given the SVD of a matrix A ∈ C^m×n

A = UΣV^H (1)

with the unitary matrices^r×r U ∈ C^m×m, V ∈ C^n×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)

• What are the dimensions of the matrices U₁, U₂, V₁, V₂, Σ 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)

2. 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.

3. 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).)

4. The system Ax = b with A ∈ C^m×n, x ∈ Cⁿ and b ∈ C^m 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. x_LS = A⁺b, is really an exact solution to the modified system AxPb
• the error of the righthand sides is orthogonal to bb.