Given what a reflection matrix does on a subspace, find the subspace - Can't solve

Question

I can't really solve this exercise I've been trying to solve for some time now. It goes like that:

Matrix $R$ ($\in \mathbb R^{3\times3}$) is a reflection matrix, in relation to subspace $U$, $U=\text{span}\{u_1\}$, $\dim U=1$. It is given that $R\left(\begin{array}{c}3\\ 3\\ 3\end{array}\right)= \left(\begin{array}{c}1\\ 1\\ 5\end{array}\right)$ . Find a basis of $U$.

So what I assumed is that I actually need to find $u_1$, and what I tried to do is to firstly find $P$. I tried to do it as follows: Projection matrix on a line $\text{span}\{u_1\}$ is given by $P=(uu^t)/(u^tu)$, then $R=2P-I$, So instead of the equation that was given, I substituted $2P-I$ and found that $P\left(\begin{array}{c}3\\ 3\\ 3\end{array}\right)=\left(\begin{array}{c}2\\ 2\\ 4\end{array}\right)$. I can't really see a way to find $u_1$..

Answer 1

I haven't checked your answer, but if what you already have is correct you are almost done: $$ P\left(\begin{array}{c}3\\ 3\\ 3\end{array}\right) = \frac{uu^T}{u^Tu} \left(\begin{array}{c}3\\ 3\\ 3\end{array}\right)=\left(\begin{array}{c}2\\ 2\\ 4\end{array}\right) \implies \underbrace{\frac{u^T}{u^Tu} \left(\begin{array}{c}3\\ 3\\ 3\end{array}\right)}_{\text{scaler}} u =\left(\begin{array}{c}2\\ 2\\ 4\end{array}\right) $$ That is you have found $u$ up to a scaler multiple.

Answer 2

Use the geometry of the situation. If we’re reflecting in a line ($\dim U=1$), then that line must be the angle bisector of $v$ and $Rv$: it’s coplanar with $v$ and $Rv$, orthogonal to $v-Rv$ and $v$ and $Rv$ have the same length. Hence, if $v\notin U$, then $U$ is spanned by $v+Rv$.

We can also reason this way: Decompose $v=v_\parallel+v_\perp$ into components in $U$ and $U^\perp$. We then have $Rv=v_\parallel-v_\perp$ and so $v+Rv=2v_\parallel\in U$. But since $U$ is one-dimensional, if $v_\parallel\ne0$ then it spans $U$, and you’ve already computed such a vector.