Linear Algebra and Its Applications, Exercise 3.3.17

Exercise 3.3.17. Find the projection matrix $P$ that projects vectors in $\mathbb{R}^2$ onto the line $x+y=0$ .

Answer: The vector $(1, -1)$ is a basis for the subspace being projected onto, which is thus the column space of

$A = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Using the formula $P = A(A^TA)^{-1}AT$ we have

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

so that $(A^TA)^{-1} = \frac{1}{2}$ and

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

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

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.

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Connecting to %s