Fixed point of translation then reflection Suppose $g$ is the map defined by translating by $(4,0)$ and then reflecting across $y=3x$.
I am wondering if $g$ fixes some point. Visually, such a point $P:=(a,b)$ would have to lie to the left of $y=3x$, and given the perpendicular line joining $P$ and $y=3x$, we would need the distance between $P$ and this point $Q$ of intersection to equal the distance between $(a+4,0)$ and $Q$. 
But how can I show if such a $P$ exists?
 A: $g\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\frac{1}{5}\begin{bmatrix}-4x+3y-16\\3x+4y+12\end{bmatrix},$ and $g\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x\\y\end{bmatrix}$ if and only if $$\begin{bmatrix}9&-3\\-3&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-16\\12\end{bmatrix}.$$ But clearly every vector on the left has its first component equal to $-3$ times its second component, and the vector on the right hand side does not have this property, thus $g$ has no fixed point.
To find the explicit form for $g,$ use a Householder transformation, which induces a reflection. Since you want to take $[3,-1]^{T}$ to $[-3,1]^{T}$ (this is the orthogonal direction to the line across which you're reflecting, which is the span of $[1,3]^{T}$) the matrix you're looking for is $$I-\frac{1}{5}\begin{bmatrix}9&-3\\-3&1\end{bmatrix}=I-2\left(\frac{1}{\sqrt{10}}\begin{bmatrix} 3\\ -1\end{bmatrix}\right)\left(\frac{1}{\sqrt{10}}\begin{bmatrix}3&-1\end{bmatrix}\right).$$ You obtain $g$ by first computing the translation, then applying this matrix.
A: Disclaimer: With such geometric linear algebra questions, I try to clearly distinguish between points (as elements of the plane) and vectors (as abstract entities and elements of a vector space). This is because points cannot be perpendicular, but vectors can. I write vectors as columns and points as tuples.
So let $\mathbb E^2 = ℝ × ℝ$ be the euclidean plane, realised as pairs of real numbers, and let $V = ℝ^{2×1}$ be the euclidean space of two-dimensional real columns.
What follows is a general principle for coming up with reflection matrices, even in higher dimensions.
The line “$y = 3x$” in $\mathbb E^2$ corresponds to the kernel of the linear map $\mathbb E^2 → \mathbb E^2, (x,y) ↦ 3x - y$. Let’s instead move to vectors where we have matrix multiplication to help with calculations.
Here the corresponding line is given by the kernel of the linear map $(v^T·\,) \colon V → V,~w ↦v^Tw$, where $v = [\begin{smallmatrix} 3 \\ - 1\end{smallmatrix}]$. The reflection on this line $L = \ker (v^T·\,)$ is the unique linear transformation $V → V$, that fixes all of $L$, and sends all of the orthogonal compliment $L^⊥$ of $L$ to their negatives.
Now, the vector $v' = [\begin{smallmatrix} 1 \\ 3 \end{smallmatrix}]$ is perpendicular to $v$, as $v^T·v' = 0$, in particular $v' ∈ L$ and $v ∈ L^⊥$. A matrix having $v$ as eigenvector and $v'$ in the kernel is $v·v^T$, as
$$(v·v^T)·v = v·(v^T·v) \quad \text{and}\quad (v·v^T)·v' = v·(v^T·v') = 0,$$
So in particular, $A = \frac{-1}{v^T·v}·v·v^T$ has $v' ∈ \ker A$ and $v$ as eigenvector with eigenvalue $-1$. Therefore,
$$B = E_2 + 2A$$
fulfills $Bv = v + 2Av = v - 2v = -v$ and $Bv' = v' + 2Av' = v'$, thus is the needed reflection on $L$.
Speficically, $v^T·v = 10$, $v·v^T = [ \begin{smallmatrix} 9 & - 3 \\ -3 & 1\end{smallmatrix} ]$, and so
$$ B = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} + \frac{-1} 5 \begin{bmatrix} 9 & - 3 \\ -3 & 1\end{bmatrix}.$$
Now, given any linear transformation $T \colon V → V,~v ↦ B(v + w)$ where $B$ is some $2×2$ matrix and $w ∈ V$, any fixed point of $T$ is a vector $v ∈ V$ with $Bv + Bw = v$, so with $Bv - v = -Bw$. So the question can be translated to whether $-Bw$ is in the image of $B - E_2$.
In our case $B$ is as above (and so $B - E_2 = 2A$) and $w = [\begin{smallmatrix} 4 \\ 0\end{smallmatrix}]$.
A: Let $g = g_2\cdot g_1$, where  $ g_1 (x, y) = (x + 4, y)$ is the translation, 
and 
$g_2$  is the reflection across the line $y = 3x.$
Consider a unit vector $\vec u$ from $(0,0)$ along $y = 3x$ in the first quadrant: 
$\vec u = (1/√10) (1,3).$
The projection of  $ (x, y)$ along $ \vec u$ :
$(x, y) \cdot (1/√10) (1, 3) = (1/√10)(x + 3y).$
$\vec d := (1/√10)(x + 3y) \vec u - (x, y).$
The reflected vector $ (x', y')$:
$(x', y') = (x, y) + 2 \vec d .$
$(x', y') = (x, y) $ 
$+ 2(1/√10)(x + 3y)(1/√10)(1, 3)$ 
$-2(x, y).$
$(x', y') = (2/10)(x + 3y)(1, 3)$ 
$- (x, y)$.
Combining:
$g = g_2 \cdot g_1 (x, y) = g_2(x +4, y)=$
$(2/10)(x + 4 + 3y)(1, 3) - (x + 4, y)$.
Is there a fixed point?
