Linear Algebra and Its Applications, Exercise 3.3.3

Exercise 3.3.3. Given

$A = \begin{bmatrix} 1&0 \\ 0&1 \\ 1&1 \end{bmatrix} \qquad b = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$

solve $Ax = b$ to find $p = A\bar{x}$ . Show that $b-p$ is orthogonal to every column in $A$ .

Answer: We have $A^TA\bar{x} = A^Tb$ or $\bar{x} = \left(A^TA\right)^{-1}A^Tb$ if $A^TA$ is invertible. In this case we have

$A^TA = \begin{bmatrix} 1&0&1 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 0&1 \\ 1&1 \end{bmatrix} = \begin{bmatrix} 2&1 \\ 1&2 \end{bmatrix}$

so that

$\left(A^TA\right)^{-1} = \frac{1}{2 \cdot 2 - 1 \cdot 1} \begin{bmatrix} 2&-1 \\ -1&2 \end{bmatrix}$

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

We then have

$\bar{x} = \begin{bmatrix} \frac{2}{3}&-\frac{1}{3} \\ -\frac{1}{3}&\frac{2}{3} \end{bmatrix} \begin{bmatrix} 1&0&1 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$

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

and

$p = A\bar{x} = \begin{bmatrix} 1&0 \\ 0&1 \\ 1&1 \end{bmatrix} \begin{bmatrix} \frac{1}{3} \\ \frac{1}{3} \end{bmatrix} = \begin{bmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{bmatrix}$

The error vector is then

$b - p = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} - \begin{bmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \\ \frac{2}{3} \\ -\frac{2}{3} \end{bmatrix}$

The inner product of the error vector $b-p$ with column 1 of $A$ is

$\begin{bmatrix} \frac{2}{3}&\frac{2}{3}&-\frac{2}{3} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \frac{2}{3} + 0 - \frac{2}{3} = 0$

The inner product of $b-p$ with column 2 of $A$ is

$\begin{bmatrix} \frac{2}{3}&\frac{2}{3}&-\frac{2}{3} \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = 0 + \frac{2}{3} - \frac{2}{3} = 0$

Thus $b-p$ is othogonal to all columns of $A$ (and thus to the column space of $A$ as well).

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, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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Linear Algebra and Its Applications, Exercise 3.3.3

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