Linear Algebra and Its Applications, Exercise 3.4.12

Exercise 3.4.12. Given the vectors a_1 = (1, 1) and a_2 = (4, 0), find a scalar c such that a_2 - ca_1 is orthogonal to a_1. Given the matrix A = \begin{bmatrix} 1&4 \\ 1&0 \end{bmatrix} whose columns are a_1 and a_2 respectively, find matrices Q and R such that Q is orthogonal and A = QR.

Answer: We must have a_1^T(a_2 - ca_1) = 0. This implies that a_1^Ta_2 = c(a_1^Ta_1) or

c = \frac{a_1^Ta_2}{a_1^Ta_1} = \frac{1 \cdot 4 + 1 \cdot 0}{1 \cdot 1 + 1 \cdot 1} = \frac{4}{2} = 2

So the scalar multiplying a_1 is 2.

As a check, we then have

a_2 - ca_1 = \begin{bmatrix} 4 \\ 0 \end{bmatrix} - 2 \cdot \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix}

with a_1^T(a_2 - ca_1) = 1 \cdot 2 + 1 \cdot -2 = 2 - 2 = 0. So the new vector is orthogonal to a_1.

We now attempt to factor A = \begin{bmatrix} 1&4 \\ 1&0 \end{bmatrix} into QR. If we perform Gram-Schmidt orthogonalization on A, the first column of Q is q_1 = a_1 / \|a_1\|. We have \|a_1\| = \sqrt{a_1^Ta_1} = \sqrt{2}, so q_1 = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}). The length \|a_1\| = \sqrt{2} would then be the 1, 1 entry of the matrix R. We thus have

Q = \begin{bmatrix} \frac{1}{\sqrt{2}}&? \\ \frac{1}{\sqrt{2}}&? \end{bmatrix} \qquad R = \begin{bmatrix} \sqrt{2}&? \\ ?&? \end{bmatrix}

The second column q_2 of Q is created by first subtracting from a_2 the projection of a_2 onto q_1 and then normalizing the result. The result of subtracting the projection is

a_2 - (q_1^Ta_2)q_1 = \begin{bmatrix} 4 \\ 0 \end{bmatrix} - (\frac{1}{\sqrt{2}} \cdot 4 + \frac{1}{\sqrt{2}} \cdot 0) \cdot \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}

= \begin{bmatrix} 4 \\ 0 \end{bmatrix} - \frac{4}{\sqrt{2}} \cdot \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \end{bmatrix} - \begin{bmatrix} 2 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix}

The length of this vector is \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2 \sqrt{2}, so we have

q_2 = \frac{1}{2 \sqrt{2}} \begin{bmatrix} 2 \\ -2 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{bmatrix}

The matrix Q is then

Q = \begin{bmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \end{bmatrix}

The second diagonal entry of the matrix R is the length 2\sqrt{2} used in computing q_2. The off-diagonal element of R is the value q_1^Ta_2 = \frac{4}{\sqrt{2}} = 2 \sqrt{2} used in subtracting from a_2 the component in the direction of q_1. We thus have

R = \begin{bmatrix} \sqrt{2}&2\sqrt{2} \\ 0&2\sqrt{2} \end{bmatrix}

The product of the two matrices is then

QR = \begin{bmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \sqrt{2}&2\sqrt{2} \\ 0&2\sqrt{2} \end{bmatrix} = \begin{bmatrix} 1&4 \\ 1&0 \end{bmatrix} = A

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.

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