Linear Algebra and Its Applications, Exercise 3.4.27

Exercise 3.4.27. Given the subspace spanned by the three vectors

$a_1 = \begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \end{bmatrix} \qquad a_2 = \begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix} \qquad a_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ -1 \end{bmatrix}$

find vectors $q_1$, $q_2$, and $q_3$ that form an orthonormal basis for the subspace.

Answer: We can save some time by noting that $a_1$ and $a_3$ are already orthogonal. We can normalize these two vectors to create $q_1$ and $q_3$:

$\|a_1\|^2 = 1^2 + (-1)^2 + 0^2 + 0^2 = 1 + 1 = 2$

$q_1 = a_1/\|a_1\| = \frac{1}{\sqrt{2}} a_1 = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \\ 0 \end{bmatrix}$

$\|a_3\|^2 = 0^2 + 0^2 + 1^2 + (-1)^2 = 1 + 1 = 2$

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

We can then compute a third orthogonal vector $a_2'$ by subtracting from $a_2$ its projections on $q_1$ and $q_3$:

$a_2' = a_2 - (q_1^Ta_2)q_1 - (q_3^Ta_2)q_3$

$= a_2 - \left[ \frac{1}{\sqrt{2}} \cdot 0 + (-\frac{1}{\sqrt{2}}) \cdot 1 + 0 \cdot (-1) + 0 \cdot 0 \right]q_1 - \left[ 0 \cdot 0 + 0 \cdot 1 + \frac{1}{\sqrt{2}} \cdot (-1) + (-\frac{1}{\sqrt{2}}) \cdot 0 \right]q_3$

$= a_2 - (-\frac{1}{\sqrt{2}})q_1 - (-\frac{1}{\sqrt{2}})q_3 = a_2 + \frac{1}{\sqrt{2}}q_1 + \frac{1}{\sqrt{2}}q_3$

$= \begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix} + \frac{1}{\sqrt{2}} \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \\ 0 \end{bmatrix} + \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \\ 0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix} + \begin{bmatrix} \frac{1}{2} \\ -\frac{1}{2} \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \frac{1}{2} \\ -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \\ -\frac{1}{2} \end{bmatrix}$

Finally, we normalize $a_2'$ to create $q_2$:

$\|a_2'\|^2 = (\frac{1}{2})^2 + (\frac{1}{2})^2 + (-\frac{1}{2})^2 + (-\frac{1}{2})^2 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1$

$q_2 = a_2'/\|a_2'\| = a_2' = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \\ -\frac{1}{2} \end{bmatrix}$

An orthonormal basis for the space is therefore

$q_1 = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \\ 0 \end{bmatrix} \qquad q_2 = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \\ -\frac{1}{2} \end{bmatrix} \qquad q_3 = \begin{bmatrix} 0 \\ 0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{bmatrix}$

(It’s worth noting that the solution for this exercise on page 480 is different than the solution given above. That’s presumably because we computed the orthonormal vectors in the order $q_1$, $q_3$, $q_2$ rather than the standard order $q_1$, $q_2$, $q_3$, taking advantage of the fact that the original vectors $a_1$ and $a_3$ were already orthogonal. Recall that a basis set is not unique, so it is possible to have different orthonormal bases for the same subspace.)

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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