Linear Algebra and Its Applications, Exercise 3.4.15

Exercise 3.4.15. Given the matrix

$A = \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix}$

find the orthonormal vectors $q_1$ and $q_2$ that span the column space of $A$ . Next find the vector $q_3$ that completes the orthonormal set, and describe the subspace of $A$ of which $q_3$ is an element. Finally, for $b = (1, 2, 7)$ find the least squares solution $\bar{x}$ to $Ax = b$ .

Answer: With $a$ and $b$ as the columns of $A$ , we first choose $a' = a = (1, 2, -2)$ . We then have

$b' = b - \frac{(a')^Tb}{(a')^Ta'}a' = b - \frac{1 \cdot 1 + 2 \cdot (-1) + (-2) \cdot 4}{1^2 + 2^2 + (-2)^2}a' = b - \frac{-9}{9}a' = b + a'$

$= \begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix} + \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix}$

Now that we have calculated the orthogonal vectors $a'$ and $b'$ we can normalize them to create the orthonormal vectors $q_1$ and $q_2$ . We have

$\|a'\| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$

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

so that

$q_1 = a' / \|a'\| = \frac{1}{3} \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} \\ \frac{2}{3} \\ -\frac{2}{3} \end{bmatrix}$

$q_2 = b' / \|b'\| = \frac{1}{3} \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{bmatrix}$

Since $a$ and $b$ span the column space of $A$ , and the orthonormal vectors $q_1$ and $q_2$ are linear combinations of $a$ and $b$ , $q_1$ and $q_2$ also span the column space of $A$ .

Next, we calculate $q_3$ . We can do this by orthogonalizing any vector $c$ that is linearly independent of $a$ and $b$ . For ease of calculation we start with $c = (1, 0, 0)$ . We then have

$c' = c - \frac{(a')^Tc}{(a')^Ta'}a' - \frac{(b')^Tc}{(b')^Tb'}b' = c - \frac{1 \cdot 1 + 2 \cdot 0 + (-2) \cdot 0}{1^2 + 2^2 + (-2)^2}a' - \frac{2 \cdot 1 + 1 \cdot 0 + 2 \cdot 0}{1^2 + 2^2 + (-2)^2}b' = c - \frac{1}{9}a' - \frac{2}{9}b'$

$= \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} - \frac{1}{9} \begin{bmatrix} 1 \\ 2 \\ -2 \end{bmatrix} - \frac{2}{9} \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} \frac{4}{9} \\ -\frac{4}{9} \\ -\frac{2}{9} \end{bmatrix}$

To normalize $c'$ we divide by

$\|c'\| = \sqrt{(-\frac{4}{9})^2 + (-\frac{4}{9})^2 + (-\frac{2}{9})^2} = \sqrt{\frac{36}{81}} = \frac{2}{3}$

so that

$q_3 = c' / \|c'\| = \frac{3}{2} \begin{bmatrix} \frac{4}{9} \\ -\frac{4}{9} \\ -\frac{2}{9} \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \\ -\frac{2}{3} \\ -\frac{1}{3} \end{bmatrix}$

Of the four fundamental subspaces of $A$ , the left nullspace $\mathcal{N}(A^T)$ is orthogonal to the column space $\mathcal{R}(A)$ . Since $q_1$ and $q_2$ span the column space $\mathcal{R}(A)$ and $q_3$ is orthogonal to $q_1$ and $q_2$ , $q_3$ must be an element of the left nullspace $\mathcal{N}(A^T)$ .

Finally, to find the least squares solution $\bar{x}$ to $Ax = b$ where $b = (1, 2, 7)$ , we factor $A = QR$ and take advantage of the fact that $R\bar{x} = Q^Tb$ .

The matrix $Q$ is simply the 3 by 2 matrix with columns $q_1$ and $q_2$ :

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

The upper triangular matrix $R$ is a 2 by 2 matrix with

$R = \begin{bmatrix} q_1^Ta&q_1^Tb \\ 0&q_2^Tb \end{bmatrix} = \begin{bmatrix} \frac{1}{3} \cdot 1 + \frac{2}{3} \cdot 2 + (-\frac{2}{3}) \cdot (-2)&\frac{1}{3} \cdot 1 + \frac{2}{3} \cdot (-1) + (-\frac{2}{3}) \cdot 4 \\ 0&\frac{2}{3} \cdot 1 + \frac{1}{3} \cdot (-1) + \frac{2}{3} \cdot 4 \end{bmatrix}$

$= \begin{bmatrix} 3&-3 \\ 0&3 \end{bmatrix}$

On the right side of the equation $R\bar{x} = Q^Tb$ we have

$Q^Tb = \begin{bmatrix} \frac{1}{3}&\frac{2}{3}&-\frac{2}{3} \\ \frac{2}{3}&\frac{1}{3}&\frac{2}{3} \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 7 \end{bmatrix} = \begin{bmatrix} -\frac{9}{3} \\ \frac{18}{3} \end{bmatrix} = \begin{bmatrix} -3 \\ 6 \end{bmatrix}$

so that the entire system is then

$\begin{bmatrix} 3&-3 \\ 0&3 \end{bmatrix} \begin{bmatrix} \bar{x_1} \\ \bar{x_2} \end{bmatrix} = \begin{bmatrix} -3 \\ 6 \end{bmatrix}$

From the second equation we have $3\bar{x_2} = 6$ or $\bar{x_2} = 2$ . Substituting into the first equation we have $3\bar{x_1} - 3\bar{x_2} = 3\bar{x_1} - 3 \cdot 2 = 3$ so that $3\bar{x_1} = 3 + 6 = 9$ or $\bar{x_1} = 3$ .

The least squares solution to $Ax = b$ with $b = (1, 2, 7)$ is therefore $\bar{x} = (3, 2)$ .

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.

Linear Algebra and Its Applications, Exercise 3.4.15

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