# Thread: projection on a plane

1. ## projection on a plane

I'm preparing for a test and somewhat noticed I don't understand everything as well as I thought I did.

I'm trying to solve the following problem:

Let $\sigma:\mathbb{R}^3\to \mathbb{R}^3$ be the projection on the plane $V: x-2y+z=0$

(a) Give a basis of $\mathbb{R}^3$ consisting of eigenvectors of $\sigma$

Since $(\sigma-I)v = 0$ for all v in V, I guess we can take two orthogonal vectors $v_1,v_2\in V$. But how do we get a third eigenvector? Just taking a third orthogonal vector?

(b) Give the matrix of $\sigma$ with respect to the standard-basis in $\mathbb{R}^3$

(c) Is $\sigma$ normal?
So since $\sigma$ is real I should figure out whether $\sigma\sigma^T = \sigma^T\sigma$. I guess I should find $\sigma$ first.

Any help is appreciated.

2. Originally Posted by Dinkydoe
I'm preparing for a test and somewhat noticed I don't understand everything as well as I thought I did.

I'm trying to solve the following problem:

Let $\sigma:\mathbb{R}^3\to \mathbb{R}^3$ be the projection on the plane $V: x-2y+z=0$

(a) Give a basis of $\mathbb{R}^3$ consisting of eigenvectors of $\sigma$

Since $(\sigma-I)v = 0$ for all v in V, I guess we can take two orthogonal vectors $v_1,v_2\in V$. But how do we get a third eigenvector? Just taking a third orthogonal vector?

(b) Give the matrix of $\sigma$ with respect to the standard-basis in $\mathbb{R}^3$

(c) Is $\sigma$ normal?
So since $\sigma$ is real I should figure out whether $\sigma\sigma^T = \sigma^T\sigma$. I guess I should find $\sigma$ first.

Any help is appreciated.

(1) Find an orthonormal basis for the plane $x-2y+z=0$ , say $\{u_1,u_2\}$ (use here Gram-Schmidt with any basis of the plane)

(2) For any $v=\begin{pmatrix}x\\y\\z \end{pmatrix} \in\mathbb{R}^3$, its (orthogonal) projection on the plane is $\sigma(v):=u_1+u_2$ , with $<,>$ the standard euclidean inner product in $\mathbb{R}^3$

(3) From the above get the matrix for $\sigma$ wrt the standard basis of $\mathbb{R}^3$ and check that indeed $\sigma^2=\sigma$

Tonio