MA514 - HW 3 - Solved

Exercise 6.1
If P is an orthogonal projector, then I − 2P is unitary. Prove this algebraically, and given a geometric interpretation.

Answer: Suppose P is an orthogonal projector. Since P is a projector, P2 = P by definition. By Theorem 6.1, P is orthogonal if and only if P = P∗. To show that I−2P is unitary, we need to show (I−2P)∗(I−2P) = (I − 2P)(I − 2P)∗ = I.

(I − 2P)∗(I − 2P) = (I∗ − 2P∗)(I − 2P) = (I − 2P)(I − 2P)

= I2 − 4P + 4P2 = I − 4P + 4P = I

(I − 2P)(I − 2P)∗ = (I − 2P)(I∗ − 2P∗) = (I − 2P)(I − 2P)

= I2 − 4P + 4P2 = I − 4P + 4P = I

Exercise 6.2
Let E be the m × m matrix that extracts the ”even part” of an m-vector: Ex = (x + Fx)/2, where F is the m × m matrix that flips (x1,...,xm)∗ to (xm,...,x1)∗. Is E is an orthogonal projector, an oblique projector, or not a projector at all? What are the entries of E?

Answer: First note that for any x = (x1,...,xm)∗ ∈ Cm, we have F2x = F(Fx) = F(F(x1,...,xm)∗) = F(xm,...,x1)∗ = (x1,...,xm)∗. This shows that F2 = I. To see if E is a projector, we check whether E2 = E to meet the definition.

 

Therefore, we know at least that E is a projector. To see if E is orthogonal, we check whether E∗ = E (according to Theorem 6.1). First note that for F to perform the transformation as the problem statement describes, it must be that F is the m × m matrix with ones on the antidiagonal and zeros elsewhere. That is,

                     0     0     ...     0     1

                     0     0     ...     1     0

           F = :      :      ...     :       : so that               F(x1,...,xm)∗ = (xm,...,x1)∗ .

                                                     

                     0     1     ...     0     0

                        1    0     ...     0    0

But then since all elements of F are real and the elements on the antidiagonal of F are all the same, F∗ = F. Using this fact,

 

Therefore, we see that E is an orthogonal projector. Since we defined an oblique projector to be a projector that is not an orthogonal projector, we conclude that E is not an oblique projector.

The entries of E are such that there are 1’s along the main diagonal and 1’s along the antidiagonal. If E is n × n for n an odd integer, however, these 1’s intersect in the very middle entry of the matrix and so in this case there is a 2 in the middlemost entry of E.

6.4
Consider the matrices

  .

(a)               What is the orthogonal projector P onto range(A), and what is the image under P of the vector (1,2,3)∗.

Answer: It is immediate that the columns of A span range(A) and since the columns of A are linearly independent, the reasoning on page 46 applies so that we can use equation (6.13) to find P. Then,

 

(b)               What is the orthogonal projector P onto range(B), and what is the image under P of the vector (1,2,3)∗.

Answer: Just as before, since the columns of B are a basis for range(B) we may apply equation (6.13). Then,

 

Exercise 7.1
Consider again the matrices A and B of Exercise 6.4.

(a)               Using any method you like, determine (on paper) a reduced QR factorization A = QˆRˆ and a full QR factorization A = QR.

Answer: Denote the columns of A by a1 and a2.

 

To get a full QR factorization we need to add a third column q3 to Qˆ such that {q1,q2,q3} are orthonormal and then add a row of zeros to Rˆ. By inspection,  will work. Therefore, a full QR factorization is:

  .

(b)               Again using any method you like, determine reduced and full QR factorizations B = QˆRˆ and B = QR.

Answer: Denote the columns of B by b1 and b2. Since b1 = a1, q1 and r11 are the same as in part (a). Next,

 

So we have the reduced QR factorization of B given by:

 

To find a full QR factorization, we find that any multiple of (−1,2,1)∗ will be orthogonal to q1 and q2. So normalizing this vector gives an option for q3 so that:

  .