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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 $u_1$..

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  • $\begingroup$ Hint: If $R$ is a reflection, then for any vector $v$, $Rv-v$ is perpendicular to the reflection (hyper-)plane. $\endgroup$ May 28, 2020 at 11:59
  • $\begingroup$ 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? $\endgroup$ May 28, 2020 at 12:00
  • $\begingroup$ @MichaelBurr r is a reflection of x on U if upholds: x+r is in U, x-r is in U perpendicular. $\endgroup$ May 28, 2020 at 12:26

2 Answers 2

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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.

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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.

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  • $\begingroup$ I don't understand how you moved the vector u to the right.. Why is it commutative? Now when you got to this point, I understand it's a scalar, but how can I find it's value without knowing u? $\endgroup$ May 28, 2020 at 12:28
  • $\begingroup$ @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 $u$. $\endgroup$
    – stochastic
    May 28, 2020 at 13:13
  • $\begingroup$ I understood now, the only problem that I have now is the fact that I need to find u, but at the same time, I am using the coordinates of u in a scalar product whose value I don't know because of u? I'll try to note u=(u1 u2 u3) and see what happens. $\endgroup$ May 28, 2020 at 16:25
  • $\begingroup$ @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. $\endgroup$
    – stochastic
    May 29, 2020 at 0:27
  • $\begingroup$ @סמיזלדין Also, if you find the answer satisfactory, I would appreciate it if you could accept the answer by clicking the checkmark next to the answer $\endgroup$
    – stochastic
    May 29, 2020 at 20:38

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