Linear Algebra and Its Applications, Exercise 3.4.6

Exercise 3.4.6. Given the matrix

$Q = \begin{bmatrix} \frac{1}{\sqrt{3}}&\frac{1}{\sqrt{14}}&\qquad \\ \frac{1}{\sqrt{3}}&\frac{2}{\sqrt{14}}&\qquad \\ \frac{1}{\sqrt{3}}&-\frac{3}{\sqrt{14}}&\qquad \end{bmatrix}$

find entries for the third column such that $Q$ is orthogonal. How much freedom do you have to choose the entries? Finally, verify that both the columns and rows are orthonormal.

Answer: In order for $Q$ to be orthogonal all three of its rows must be orthonormal, with length of 1. We start with the vector forming the first row of $Q$ . For its length to be 1 the square of the last entry of the first row must be equal to

$1 - (\frac{1}{\sqrt{3}})^2 - (\frac{1}{\sqrt{14}})^2 = 1 - \frac{1}{3} - \frac{1}{14} = 1 - \frac{14}{42} - \frac{3}{42} = \frac{25}{42}$

so that the entry itself would be $\pm \sqrt{\frac{25}{42}} = \pm \frac{5}{\sqrt{42}}$ .

Similarly the square of the last entry of the second row must be

$1 - (\frac{1}{\sqrt{3}})^2 - (\frac{2}{\sqrt{14}})^2 = 1 - \frac{1}{3} - \frac{4}{14} = 1 - \frac{14}{42} - \frac{12}{42} = \frac{16}{42}$

so that the entry itself would be $\pm \sqrt{\frac{16}{42}} = \pm \frac{4}{\sqrt{42}}$ .

Finally, the square of the last entry of the third row must be

$1 - (\frac{1}{\sqrt{3}})^2 - (\frac{3}{\sqrt{14}})^2 = 1 - \frac{1}{3} - \frac{9}{14} = 1 - \frac{14}{42} - \frac{27}{42} = \frac{1}{42}$

so that the entry itself would be $\pm \sqrt{\frac{1}{42}} = \pm \frac{1}{\sqrt{42}}$ .

We have two possible choices for the last entry of the first row, $\frac{5}{\sqrt{42}}$ and $-\frac{5}{\sqrt{42}}$ . Suppose that we choose $\frac{5}{\sqrt{42}}$ . Since the first row and the second row are supposed to be orthogonal, we must choose $-\frac{4}{\sqrt{42}}$ for the last entry of the second row, so that the dot product of the two rows is zero:

$\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{14}} \cdot \frac{2}{\sqrt{14}} + \frac{5}{\sqrt{42}} \cdot (-\frac{4}{\sqrt{42}}) = \frac{1}{3} + \frac{2}{14} - \frac{20}{42} = \frac{14}{42} + \frac{6}{42} - \frac{20}{42} = 0$

We must then choose $-\frac{1}{\sqrt{42}}$ for the last entry in the third row, so that the dot product of the first row and the third row is zero:

$\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{14}} \cdot (-\frac{3}{\sqrt{14}}) + \frac{5}{\sqrt{42}} \cdot (-\frac{1}{\sqrt{42}}) = \frac{1}{3} - \frac{3}{14} - \frac{5}{42} = \frac{14}{42} - \frac{9}{42} - \frac{5}{42} = 0$

and the dot product of the second and third rows is zero:

$\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{14}} \cdot (-\frac{3}{\sqrt{14}}) + (-\frac{4}{\sqrt{42}}) \cdot (-\frac{1}{\sqrt{42}}) = \frac{1}{3} - \frac{6}{14} + \frac{4}{42} = \frac{14}{42} - \frac{18}{42} + \frac{4}{42} = 0$

The resulting value for the matrix $Q$ is

$Q = \begin{bmatrix} \frac{1}{\sqrt{3}}&\frac{1}{\sqrt{14}}&\frac{5}{\sqrt{42}} \\ \frac{1}{\sqrt{3}}&\frac{2}{\sqrt{14}}&-\frac{4}{\sqrt{42}} \\ \frac{1}{\sqrt{3}}&-\frac{3}{\sqrt{14}}&-\frac{1}{\sqrt{42}} \end{bmatrix}$

The dot product of the first column with itself is

$(\frac{1}{\sqrt{3}})^2 + (\frac{1}{\sqrt{3}})^2 + (\frac{1}{\sqrt{3}})^2 = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$

The dot product of the second column with itself is

$(\frac{1}{\sqrt{14}})^2 + (\frac{2}{\sqrt{14}})^2 + (-\frac{3}{\sqrt{14}})^2 = \frac{1}{14} + \frac{4}{14} + \frac{9}{14} = 1$

The dot product of the third column with itself is

$(\frac{5}{\sqrt{42}})^2 + (-\frac{4}{\sqrt{42}})^2 + (-\frac{1}{\sqrt{42}})^2 = \frac{25}{42} + \frac{16}{42} + \frac{1}{42} = 1$

Thus all three columns have length 1.

We also have the dot product of the first and second columns as zero:

$\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{14}} + \frac{1}{\sqrt{3}} \cdot \frac{2}{\sqrt{14}} + \frac{1}{\sqrt{3}} \cdot (-\frac{3}{\sqrt{14}}) = \frac{1}{\sqrt{42}} + \frac{2}{\sqrt{42}} - \frac{3}{\sqrt{42}} = 0$

the dot product of the first and third columns as zero:

$\frac{1}{\sqrt{3}} \cdot \frac{5}{\sqrt{42}} + \frac{1}{\sqrt{3}} \cdot (-\frac{4}{\sqrt{42}}) + \frac{1}{\sqrt{3}} \cdot (-\frac{1}{\sqrt{42}}) = \frac{5}{\sqrt{126}} - \frac{4}{\sqrt{126}} - \frac{1}{\sqrt{126}} = 0$

and the dot product of the second and third columns as zero:

$\frac{1}{\sqrt{14}} \cdot \frac{5}{\sqrt{42}} + \frac{2}{\sqrt{14}} \cdot (-\frac{4}{\sqrt{42}}) + (-\frac{3}{\sqrt{14}}) \cdot (-\frac{1}{\sqrt{42}}) = \frac{5}{\sqrt{588}} - \frac{8}{\sqrt{588}} + \frac{3}{\sqrt{588}} = 0$

Since all three rows are orthonormal (by construction) and all three columns are orthonormal, the matrix $Q$ is an orthogonal matrix.

Recall that we originally had two choices for the last entry of the first row. If we instead choose $-\frac{5}{\sqrt{42}}$ for the last entry in the first row, we must choose $\frac{4}{\sqrt{42}}$ for the last entry of the second row and $\frac{1}{\sqrt{42}}$ for the last entry in the third row, so that

$Q = \begin{bmatrix} \frac{1}{\sqrt{3}}&\frac{1}{\sqrt{14}}&-\frac{5}{\sqrt{42}} \\ \frac{1}{\sqrt{3}}&\frac{2}{\sqrt{14}}&\frac{4}{\sqrt{42}} \\ \frac{1}{\sqrt{3}}&-\frac{3}{\sqrt{14}}&\frac{1}{\sqrt{42}} \end{bmatrix}$

Verifying that this alternative value for $Q$ is an orthogonal matrix is left as an exercise for the reader.

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 3.4.6

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