## 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

1. jay kim says:

This is great. Thank you so much.