## 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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