Linear Algebra and Its Applications, Exercise 3.2.14

Exercise 3.2.14. In $\mathbb{R}^3$ the planes corresponding to the equations $x+y+t =0$ and $x-t=0$ intersect in a line. What is the projection matrix $P$ that projects points in $\mathbb{R}^3$ onto that line?

Answer: To find the line of intersection we solve the system of equations $Av = 0$ where

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

We use Gaussian elimination, subtracting 1 times row 1 from row 2 to produce the echelon matrix

$U = \begin{bmatrix} 1&1&1 \\ 0&-1&-2 \end{bmatrix}$

Since $U$ has pivots in columns 1 and 2 we have $x$ and $y$ as basic variables and $t$ as a free variable. Setting $t=1$ from the second row of $U$ we have $-y-2t = -y-2 =0$ or $y = -2$ . From the first row of $U$ we then have $x+y+t = x -2 +1 = 0$ or $x = 1$ . So $\left(1, -2, 1 \right)$ is a solution to the system, as is any other vector on the line through the origin and $\left(1, -2, 1 \right)$ .

The projection matrix projecting points in $\mathbb{R}^3$ onto the line through $v = \left(1, -2, 1 \right)$ is then

$P = vv^T/v^Tv = \frac{1}{1 + 4 +1} \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \begin{bmatrix} 1&-2&-1 \end{bmatrix}$

$= \frac{1}{6} \begin{bmatrix} 1&-2&1 \\ -2&4&-2 \\ 1&-2&-1 \end{bmatrix} = \begin{bmatrix} \frac{1}{6}&-\frac{1}{3}&\frac{1}{6} \\ -\frac{1}{3}&\frac{2}{3}&-\frac{1}{3} \\ \frac{1}{6}&-\frac{1}{3}&\frac{1}{6} \end{bmatrix}$

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

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1 Response to Linear Algebra and Its Applications, Exercise 3.2.14

jay kim says:

September 11, 2014 at 6:38 am

This is great. Thank you so much.