The Meaning of the Orthogonal Projection Transformation

Question

Let $T: \Bbb R^3 \to \Bbb R^3$ the Orthogonal Projection Transformation on the plane $x+2y+5z=0$.

Find 5 non-trivial T-Invariant sub spaces of $\Bbb R^3$.

I believe I have to find a polynomial representation of this transformation.

All I know about the orthogonal projection is that if I had a polynomial $P=x+y$ then $T(P)=x$. However I don't know how to use it here with 3 vectors.

I do think that transformation "takes" 2 of the 3 vectors, so the final basis has to be of dimension 3.

Thank you in advance.

Answer 1

An arbitrary vector in the plane is $(x,y,-x-5y)-(0,0,0)$, so we get the subspaces

$\left \{ x(1,0,-1) +y(0,1,-5)\right \}$.

A vector normal to these subspaces is $(1,2,5)$ so that if $\vec v\in \mathbb R^3$, then

$\vec v=x(1,0,-1) +y(0,1,-5)+t(1,2,5)$ and so

$T(\vec v)=x(1,0,-1) +y(0,1,-5)$.

Answer 2

Take subspaces of the plane, for example:

$$\begin{align*}&x+2y+z=0\\{}\\&(1,0,-1)+t(-1,1,-1)\;t\in\Bbb R\\{}\\&(1,0,-1)+t(0,1,-2)\;,\;\;t\in\Bbb R\;\;,\;\;\;etc.\end{align*}$$

Answer 3

Let $P$ be the plane, then we must have $Tx = x$ for $x \in P$. Hence any subspace of $P$ is $T$ invariant. Since $\dim P = 2$, you can choose any line in $P$ through the origin.

For example, note that $(2,-1,0)$ and $(0,5,-2)$ are in $P$. Then $v_{\alpha,\beta}=\alpha (2,-1,0) + \beta (0,5,-2) \in P$ and so $T v_{\alpha,\beta} = v_{\alpha,\beta}$, and so $\operatorname{sp} \{ v_{\alpha,\beta} \} $ is a $T$ invariant subspace.

Now pick 5 values of $\alpha,\beta$ so that $v_{\alpha,\beta}$ lie on different lines.

Note that $P$ is also $T$ invariant.