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

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 $$u_1$$..

• Hint: If $R$ is a reflection, then for any vector $v$, $Rv-v$ is perpendicular to the reflection (hyper-)plane. May 28, 2020 at 11:59
• I am a bit confused about your definition of $U$. Often, a reflection in three dimensions is a reflection over a plane, so I would expect that $U$ would be two dimensional and not one-dimensional. How is your reflection defined? May 28, 2020 at 12:00
• @MichaelBurr r is a reflection of x on U if upholds: x+r is in U, x-r is in U perpendicular. May 28, 2020 at 12:26

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.
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.
• @סמיזלדין you can keep it on the left, then the right side is still a scaler and commutes with the vector. Note that multiplication is associative, so all I'm doing is to multiply $u^T$ by $(3,3,3)^T$ first, then I have a scaler and can move it passed $u$. May 28, 2020 at 13:13
• @סמיזלדין $u = (2,2,4)^T$ is your solution. You're just trying to find a basis, you don't need to know the scaler multiple. Note that if you multiply $u$ by any constant $c$, it gets canceled out in the equation, because you have the same number of $u$s in the numerator and the denominator. May 29, 2020 at 0:27