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 I2P is unitary, we need to show (I2P)(I2P) = (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,,x_m) to (x_m,,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,,x_m) C^m, we have F²x = F(Fx) = F(F(x₁,,x_m)) = F(x_m,,x₁) = (x₁,,x_m). 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,,x_m) = (x_m,,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 1s along the main diagonal and 1s along the antidiagonal. If E is n n for n an odd integer, however, these 1s 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

.

• 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,

• 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.

• Using any method you like, determine (on paper) a reduced QR factorization A = QR 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 Q such that {q₁,q₂,q₃} are orthonormal and then add a row of zeros to R. By inspection, will work. Therefore, a full QR factorization is:

.

• Again using any method you like, determine reduced and full QR factorizations B = QR 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:

.