# Matrix of the projection

Find the matrix of the projection in $\mathbb{R}^3$ onto the plane $x_1=x_3.$

I can find a normal unit vector of the plane, which is $\vec{n}=(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}})^T$

And the projection is $$\text{proj}_V(\vec{x})=\vec{x}-(\vec{x}\dot\ \vec{n})\vec{n},$$ but why would the solution be $$A = \begin{pmatrix} \frac12 & 0 & \frac12 \\ 0 & 1 & 0 \\ \frac12 & 0 & \frac12 \end{pmatrix}?$$

• $\vec n$ is not normal to the plane, it is in the plane. Feb 8, 2015 at 10:00
• @MatthewLeingang Ops, I fixed it. Thanks! Feb 8, 2015 at 10:02

## 2 Answers

this is basically a two dimensional projection because $e_2 = (0,1,0)^T$ is on the plane. so you have projection on to the diagonal line $x_1 = x_3$ in the $x_1x_3$ plane. both $(1,0,0)^T$ and $(0,0,1)^T$ project to $(1/2, 0, 1/2),$ the midpoint of the hypotenuse of an isosceles right angle triangle of unit sides. now you can set up the projection matrix $P$ by getting the columns $Pe_1 = (1/2, 0, 1/2) = Pe_3, Pe_2 = e_2.$

Hint: Recall that for example the first column of $A$ is given by $Ae_1$. So

$$Ae_1 = e_1 - (e_1 \cdot \vec n) \vec n = (1,0,0)^T - \frac{1}{\sqrt 2} (\frac{1}{\sqrt 2} , 0, -\frac{1}{\sqrt 2})^T = (1,0,0)^T + (-\frac 12 , 0, \frac 12)^T$$

• That's the "canonical" way to represent a lienar operator by a matrix. @Robben
– user99914
Feb 8, 2015 at 10:50