Let $T : \mathbb R^2 → \mathbb R^2$ the orthogonal projection on the line $y = mx$. Prove that for all $a, b \in \mathbb R$,
$$\begin{align}T((a,b)) = {\frac{1}{m^2 + 1}}(a+mb, ma + m^2b)\end{align}$$
The mark scheme talks about using an equation $$\begin{align}y = {\frac{-1}{m}x + b + \frac{a}{m}}\end{align}$$
First of all where did they get that equation from? I've never heard of it and it hasnt been given to us in our lecture notes. And where do i go from there to find the answer above.
Then it says let : $$\begin{align}S:\mathbb R^2 → \mathbb R^2 \end{align}$$ be the orthogonal symmetry with respect to the line $y = mx$. Prove that for all $a, b \in \mathbb R$,
$$\begin{align} S((a, b)) = −(a, b) + 2T ((a, b)) \end{align}$$ So I can visualise it but I dont understand how the mark scheme can write: $v \in \mathbb R^2$ and let $x \in \mathbb R^2$ such that $T(v) + x = v$. then $S(v) = T (v) − x = T (v) − (v − T (v)) = 2T (v) − v.$ Because I dont get how $T(v) + x = v$? it wouldnt it equal $x$ not $v$? I am sure the mark scheme is right but i cant seem to understand that part. All i know is it would be $2$ times the projection minus the vector $(a,b)$.
Here's how I solve this problem:
Notice I am writing vectors in columnar form; thus, the OP's $(a, b)$ is my
$\begin{pmatrix} a \\ b \end{pmatrix} \tag{0}$
etc.; these things being said , we have:
The line $y = mx$ is in fact the set of points (vectors) $L = \{(x, mx)^T = x(1, m)^T \mid x \in \Bbb R \}$; this is the one-dimensional subspace of $\Bbb R^2$ generated by the vector $\vec v = (1, m)^T$. $L$ is also generated by the unit vector $\mathbf u_{\vec v}$ in the direction of $\vec v$, and since $\Vert \vec v \Vert = \sqrt{m^2 + 1}$ we have
$\mathbf u_{\vec v} = \begin{pmatrix} \dfrac{1}{\sqrt{m^2 + 1}} \\ \dfrac{m}{\sqrt{m^2 + 1}} \end{pmatrix}. \tag{1}$
The scalar component of the projection of any vector $(a, b)^T \in \Bbb R^2$ onto the subspace $L = \langle \mathbf u_{\vec v} \rangle$ generated by the unit vector $\mathbf u_{\vec v}$ is in fact
$\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v} = \begin{pmatrix} a \\ b \end{pmatrix} \cdot \begin{pmatrix} \dfrac{1}{\sqrt{m^2 + 1}} \\ \dfrac{m}{\sqrt{m^2 + 1}} \end{pmatrix} = \dfrac{a + mb}{\sqrt{m^2 + 1}}, \tag{2}$
and the projected vector itself is this scalar component times $\mathbf u_{\vec v}$, or
$T((a, b)^T) = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v}) \mathbf u_{\vec v} = \dfrac{a + mb}{\sqrt{m^2 + 1}}\begin{pmatrix} \dfrac{1}{\sqrt{m^2 + 1}} \\ \dfrac{m}{\sqrt{m^2 + 1}} \end{pmatrix} = \begin{pmatrix} \dfrac{a + mb}{m^2 + 1} \\ \dfrac{ma + m^2 b}{m^2 + 1}\end{pmatrix}; \tag{3}$
thus is the OP's first equation established.
As for the equation for $S((a, b)^T$,
$S((a, b)^T) = -(a, b)^T + 2T((a, b)^T) = -\begin{pmatrix} a \\ b \end{pmatrix} + 2T(\begin{pmatrix} a \\ b \end{pmatrix}), \tag{4}$
note that we have
$\begin{pmatrix} a \\ b \end{pmatrix} = T(\begin{pmatrix} a \\ b \end{pmatrix}) + (\begin{pmatrix} a \\ b \end{pmatrix} - T(\begin{pmatrix} a \\ b \end{pmatrix})); \tag{5}$
furthermore, we see that
$T(\begin{pmatrix} a \\ b \end{pmatrix}) \cdot (\begin{pmatrix} a \\ b \end{pmatrix} - T(\begin{pmatrix} a \\ b \end{pmatrix}) = 0. \tag{6}$
(6) is easily seen using
$T((a, b)^T) = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v}) \mathbf u_{\vec v}, \tag{7}$
the leftmost equation in (3) and the definition of $T((a, b)^T)$; indeed
$T((a, b)^T) \cdot T((a, b)^T) = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v})^2 \Vert \mathbf u_{\vec v} \Vert^2 = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v})^2 \tag{8}$
since $\Vert \mathbf u_{\vec v} \Vert^2 = 1$, and
$T((a, b)^T) \cdot \begin{pmatrix} a \\ b \end{pmatrix} = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v}) \mathbf u_{\vec v} \cdot \begin{pmatrix} a \\ b \end{pmatrix} = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v})^2 \tag{9}$
as well, so that (6) indeed binds, showing that $(a, b)^T - T((a, b)^T)$ is orthogonal to the subspace $L = \langle \mathbf u_{\vec v} \rangle$. Now the reflection $S$ about/across the line $L$ leaves elements of $L$ itself invariant, and reverses vectors in $\Bbb R^2$ which are orthogonal to $L$; thus we must have
$ST(\begin{pmatrix} a \\ b \end{pmatrix}) = T(\begin{pmatrix} a \\ b \end{pmatrix}), \tag{10}$
but
$S(\begin{pmatrix} a \\ b \end{pmatrix} - T(\begin{pmatrix} a \\ b \end{pmatrix})) = T(\begin{pmatrix} a \\ b \end{pmatrix}) - \begin{pmatrix} a \\ b \end{pmatrix}; \tag{11}$
applying (10) and (11) to (5) yields
$S(\begin{pmatrix} a \\ b \end{pmatrix}) = ST(\begin{pmatrix} a \\ b \end{pmatrix}) + S(\begin{pmatrix} a \\ b \end{pmatrix} - T(\begin{pmatrix} a \\ b \end{pmatrix})) = T(\begin{pmatrix} a \\ b \end{pmatrix}) - \begin{pmatrix} a \\ b \end{pmatrix} + T(\begin{pmatrix} a \\ b \end{pmatrix})$ $= 2T(\begin{pmatrix} a \\ b \end{pmatrix}) - \begin{pmatrix} a \\ b \end{pmatrix}, \tag{12}$
as was to be shown. QED.
Additional Remarks added Thursday 24 July 2014 10:58 PM PST: I don't know exactly what the "mark scheme" is, but I think that introduction of the line
$y = -\dfrac{x}{m} + b + \dfrac{a}{m} \tag{13}$
is meant to provide a slightly different argument, though one geometrically equivalent to that given above. For we see that the slope of line (13) is $-(1/m)$; as such, it is perpendicular to $L$. (Recall that two lines are perpendicular if their slopes $m_1$, $m_2$ satisfy $m_1 m_2 = -1$.) Furthermore, the line (13) passes through the point $(a, b)^T$ as may be seen by setting $x = a$, $y = b$; thus (13) is the equation of the line normal to $L$ containing the point $(a, b)^T$. As such, it is related to the OP's equation $T(v) + x = v$ or my equation (5); $x = v - T(v)$ is normal to $T(v)$, which lies along the line $L: y = mx$; it ($v - T(v)$) may be thought of as running from $L$ to $v$. Though I can't, being unfamiliar with "mark scheme"(s), say exactly what the referenced one suggests, it appears to be focused the relationship between the geometry of this problem, based on the lines $y = mx$ and (13), and it's linear algebra, which uses vectors such as my $\mathbf u_{\vec v}$. Perhaps these observations can help our OP user135688 piece things together along these "lines", if I may so say; he certainly is on the inside track. But the best I can do is provide the preceding mere suggestions of an idea. End of Remarks.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!