## Linear Algebra and Its Applications, Exercise 3.4.12

Exercise 3.4.12. Given the vectors $a_1 = (1, 1)$ and $a_2 = (4, 0)$, find a scalar $c$ such that $a_2 - ca_1$ is orthogonal to $a_1$. Given the matrix $A = \begin{bmatrix} 1&4 \\ 1&0 \end{bmatrix}$ whose columns are $a_1$ and $a_2$ respectively, find matrices $Q$ and $R$ such that $Q$ is orthogonal and $A = QR$.

Answer: We must have $a_1^T(a_2 - ca_1) = 0$. This implies that $a_1^Ta_2 = c(a_1^Ta_1)$ or $c = \frac{a_1^Ta_2}{a_1^Ta_1} = \frac{1 \cdot 4 + 1 \cdot 0}{1 \cdot 1 + 1 \cdot 1} = \frac{4}{2} = 2$

So the scalar multiplying $a_1$ is 2.

As a check, we then have $a_2 - ca_1 = \begin{bmatrix} 4 \\ 0 \end{bmatrix} - 2 \cdot \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix}$

with $a_1^T(a_2 - ca_1) = 1 \cdot 2 + 1 \cdot -2 = 2 - 2 = 0$. So the new vector is orthogonal to $a_1$.

We now attempt to factor $A = \begin{bmatrix} 1&4 \\ 1&0 \end{bmatrix}$ into $QR$. If we perform Gram-Schmidt orthogonalization on $A$, the first column of $Q$ is $q_1 = a_1 / \|a_1\|$. We have $\|a_1\| = \sqrt{a_1^Ta_1} = \sqrt{2}$, so $q_1 = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. The length $\|a_1\| = \sqrt{2}$ would then be the 1, 1 entry of the matrix $R$. We thus have $Q = \begin{bmatrix} \frac{1}{\sqrt{2}}&? \\ \frac{1}{\sqrt{2}}&? \end{bmatrix} \qquad R = \begin{bmatrix} \sqrt{2}&? \\ ?&? \end{bmatrix}$

The second column $q_2$ of $Q$ is created by first subtracting from $a_2$ the projection of $a_2$ onto $q_1$ and then normalizing the result. The result of subtracting the projection is $a_2 - (q_1^Ta_2)q_1 = \begin{bmatrix} 4 \\ 0 \end{bmatrix} - (\frac{1}{\sqrt{2}} \cdot 4 + \frac{1}{\sqrt{2}} \cdot 0) \cdot \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$ $= \begin{bmatrix} 4 \\ 0 \end{bmatrix} - \frac{4}{\sqrt{2}} \cdot \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \end{bmatrix} - \begin{bmatrix} 2 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix}$

The length of this vector is $\sqrt{2^2 + (-2)^2} = \sqrt{8} = 2 \sqrt{2}$, so we have $q_2 = \frac{1}{2 \sqrt{2}} \begin{bmatrix} 2 \\ -2 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{bmatrix}$

The matrix $Q$ is then $Q = \begin{bmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \end{bmatrix}$

The second diagonal entry of the matrix $R$ is the length $2\sqrt{2}$ used in computing $q_2$. The off-diagonal element of $R$ is the value $q_1^Ta_2 = \frac{4}{\sqrt{2}} = 2 \sqrt{2}$ used in subtracting from $a_2$ the component in the direction of $q_1$. We thus have $R = \begin{bmatrix} \sqrt{2}&2\sqrt{2} \\ 0&2\sqrt{2} \end{bmatrix}$

The product of the two matrices is then $QR = \begin{bmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \sqrt{2}&2\sqrt{2} \\ 0&2\sqrt{2} \end{bmatrix} = \begin{bmatrix} 1&4 \\ 1&0 \end{bmatrix} = A$

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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