Linear Algebra and Its Applications, Exercise 3.4.16

Exercise 3.4.16. Given the matrix A whose columns are the following two vectors a_1 and a_3 [sic]:

a_1 = \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} \quad a_3 = \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}

factor A as A = QR. If there are n vectors a_j with m elements each, what are the dimensions of A, Q, and R?

Answer: With a_1 and a_3 as the two columns of A, we first choose a_1' = a_1 = (1, 2, 2). We then have

a_3' = a_3 - \frac{(a_1')^Ta_3}{(a_1')^Ta_1'}a_1' = a_3 - \frac{1 \cdot 1 + 2 \cdot 3 + 2 \cdot 1}{1^2 + 2^2  + 2^2}a_1' = a_3 - \frac{9}{9}a_1' = a_3 - a_1'

= \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix} - \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}

Now that we have calculated the orthogonal vectors a_1' and a_3' we can normalize them to create the orthonormal vectors q_1 and q_3. We have

\|a_1'\| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3

\|a_3'\| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}

so that

q_1 = a_1' / \|a_1'\| = \frac{1}{3} \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{bmatrix}

q_3 = a_3' / \|a_3'\| = \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{bmatrix}

The matrix Q is the 3 by 2 matrix with columns q_1 and q_3:

Q = \begin{bmatrix} \frac{1}{3}&0 \\ \frac{2}{3}&\frac{1}{\sqrt{2}} \\ \frac{2}{3}&-\frac{1}{\sqrt{2}} \end{bmatrix}

The matrix R is the 2 by 2 matrix calculated as follows:

R = \begin{bmatrix} q_1^Ta_1&q_1^Ta_3 \\ 0&q_3^Ta_3 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} \cdot 1 + \frac{2}{3} \cdot 2 + \frac{2}{3} \cdot 2&\frac{1}{3} \cdot 1 + \frac{2}{3} \cdot 3 + \frac{2}{3} \cdot 1 \\ 0&0 \cdot 1 + \frac{1}{\sqrt{2}} \cdot 3 + (-\frac{1}{\sqrt{2}}) \cdot 1 \end{bmatrix}

= \begin{bmatrix} \frac{9}{3}&\frac{9}{3} \\ 0&\frac{2}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} 3&3 \\ 0&\sqrt{2} \end{bmatrix}

If there are n vectors a_j with m elements each, since they form the columns of A the shape of A will be m by n. The matrix Q contains one orthonormal column for each column in A, and its orthonormal columns have the same number of elements as the columns of A, so Q is also m by n. Finally, R is a square matrix with one column for each column of Q, so it is n by n.

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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