# Linear Algebra - Finding the matrix for the transformation

Question:

Find the matrix for the transformation which first reflects across the main diagonal, then projects onto the line $$2y+\sqrt{3}x=0$$, and then reflects about the line $$\sqrt{3}y=2x$$.

Attempt to Question:

Reflection about the line $$y=x: T(x,y)=(y,x)$$, so the standard matrix for this would be the matrix

$$\begin{pmatrix}0 &1 \\ 1 &0 \end{pmatrix}.$$

However I'm not sure how to deal with equations rather than axis. I assume in the second projection, you can simplify it to $$y=\frac{-\sqrt{3}}{2} x$$. Can you then separate these into a scalar operation $$\frac{-\sqrt{3}}{2}$$ and orthogonal operation $$y=x$$? Even so, I wouldn't know how to go further than this since I only know how to do it among the axis.

Edit: $$T_1=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$$ by reflecting across main diagonal

$$2y+\sqrt{3}x=0\rightarrow y=\frac{-\sqrt{3}}{2}x$$

$$T_2=\frac{-\sqrt{3}}{2}\left[\begin{array}{cc}\cos-45&-\sin-45\\\sin-45&\cos-45\end{array}\right]\left[\begin{array}{cc}0&0\\0&1\end{array}\right]\left[\begin{array}{cc}\cos45&-\sin45\\\sin45&\cos45\end{array}\right]$$ by rotating $$45^\circ$$, projecting along the y axis, then rotating $$-45^\circ$$ (transformations are applied right to left)

$$\sqrt{3}y=2x\rightarrow y=\frac{2}{\sqrt{3}}x$$

$$T_3=\frac{2}{\sqrt{3}}\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$$ by separating fraction and then reflecting along $$y=x$$

• You got a sign wrong for the equation of the line for the second transformation. Apr 2, 2013 at 18:28
• Fixed, thanks for catching that. Apr 2, 2013 at 18:34
• I tried this method and edited the original. Can you please confirm that I did it correctly? I projected onto y axis instead because rotating 45 degrees would put it on the y axis Apr 2, 2013 at 19:02
• Trying that now, thanks. I edited my answer, I didn't realize that transformations are applied that way. Apr 2, 2013 at 19:12
• I've added my completed answer. Please let me know if it is correct. Additionally, do you know how I should format the answer? I'm not sure what format to put it in other than separating the three transformations Apr 2, 2013 at 19:26

Okay, let's start with projections. The projection matrix onto a line $a x + b y = 0$ is a linear transformation expressible by a matrix, mapping the world onto points on that line. A typical point on that line has the form $t [b ;\; -a ]$ for some $t$, as this generates $a(b t) + b (-a t) = 0$. So the unit vector pointing in the direction of that line is $\hat u = [b ;\; -a] / \sqrt{a^2 + b^2}$ and the projection of a vector $\vec v$ is $\operatorname{proj}_{\hat u}~\vec v = \hat u (\hat u \cdot \vec v)$ which we can write as a matrix:$$\operatorname{proj}_{\hat u} = \frac{1}{a^2 + b^2} \begin{bmatrix}b\\-a\end{bmatrix}\begin{bmatrix}b & -a\end{bmatrix} = \frac{1}{a^2 + b^2} \begin{bmatrix}b^2 & -ba\\-ba & a^2\end{bmatrix}.$$So that's the projection matrix.
Once you have projections onto a line, you have reflections about the line. This is because if $\operatorname{proj}_{\hat u} \vec v = \vec v_u$ then we know $\vec v = \vec v_u + \vec c$ for some vector $\vec c$, and then the reflection about that line is just $\vec v_u - \vec c$: you flip the sign of the deviation, but you do not change the projection. Some thinking gives you an explicit construction as: $$\operatorname{flip}_{\hat u} \vec v = \operatorname{proj}_{\hat u} \vec v - (\vec v - \operatorname{proj}_{\hat u} \vec v) = 2 \operatorname{proj}_{\hat u} \vec v - \vec v$$so that the actual matrix which does this is $$\operatorname{flip}_{\hat u} = 2 \operatorname{proj}_{\hat u} - I.$$
Now indeed, if we're flipping around the line $y = x$ then $a = 1, b = -1$, and our projection matrix is $\frac 12 \begin{bmatrix}1&1\\1&1\end{bmatrix}$ and indeed, doubling this and subtracting the identity matrix gives $\begin{bmatrix}0&1\\1&0\end{bmatrix}.$
Your end result is therefore probably not right, since I do not see any factors of $\frac 17,$ which comes naturally from these $(a^2 + b^2)^{-1}$ prefactors.