Help in understanding this exercise (Linear Algebra)

Question

I need some help in understanding the precise request of this exercise.

Let the vectors of the plane be identified with oriented segments exiting from a fixed point, and let's identify $\mathcal{V}^2$ with $\mathbb{R}^2$ through a system of two orthogonal unit vectors. Be $S, T: \mathcal{V}^2 \to \mathcal{V}^2$ linear applications the first one denoting the symmetry with respect to the first coordinate axis, and the second one with respect to the bisectrix of the first and third quadrant of the cartesian plane. Calculate in two different ways $S\circ T$ and $T\circ S$ (that is, through matrix multiplication and through the images of the unit vectors).

So, in layman words I can say I understood what I have to do.

I tried to sketch some passages, so I would appreciate if you tell me if I'm right or not. First of all, I think the exercise is really poorly written but still.

I interpreted the symmetryes with respect to axis and bisectrix as isometries of the type of reflections (cannot be rotation, since we did not do them).

Now the problem is: how to obtain the linear application that describes a reflextion around a certain axis or line?

For what concerns the linear application $S$ for the reflection around the $x$-axis I thought that it must hold

$$S (x, 0) = (x, 0)$$

Which means: the action of the linear application $S$ over any vector of the form $(x, 0)$ must return itself. Also:

$$S(x, y) = (x, -y)$$

which generalises the previous one, so I believe this is enough to describe $S$ (?)

How to deal with the reflection $T$ around the line $y = x$?

I thought to reason in terms of basis? The vector space $y = x$ admits the basis $\mathcal{B} = \{(1, 1)\}$, but I actually don't know what to do with this. I mean, for sure we have

$$T(x, x) = (x, x)$$

But if we take for example the point $(1, 4)$ then I expect $T(1, 4) = (4, 1)$. Though how can I pass from the intruition, assuming it's correct, to the algebraic formulation?

The linear application $S$ can be retrieved by it's action on the canonical basis:

$$S(1, 0) = (1, 0)$$ $$S(0, 1) = (0, -1)$$

Hence the associated matrix is

$$A(S) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$

Then we have

$$S(x, y) = (x, -y)$$

About $T$ instead how can I instead "legitly prove" that

$$T(1, 0) = (0, 1)$$ $$T(0, 1) = (1, 0)$$

Because after that, I can indeed complete the exercise easily.

In general, supposing I have a line $y = mx$, how can I determine the linea application that reflects a point around that line?

Answer 1

Find the reflected point with respect to some line is a really trivial task in planar analytic geometry one might be familiar in middle school.

Given a point $(a,b)$ and a line $y=mx$, set the reflected point $(p,q)$ and the midpoint lies on this line so $q+b=m(p+a)$. Also their linking line is perpendicular to the given line. So $(p-a,q-b)\cdot(1,m)=0$.

Hence you get a matrix equation $\begin{bmatrix}-m&1\\1&m \end{bmatrix}\begin{bmatrix}p\\ q \end{bmatrix} =\begin{bmatrix}m&-1\\1&m \end{bmatrix}\begin{bmatrix}a\\ b \end{bmatrix}$ which is $ \begin{bmatrix}p\\ q \end{bmatrix} =\cfrac{1}{m^2+1}\begin{bmatrix}1-m^2&2m\\2m&m^2-1 \end{bmatrix}\begin{bmatrix}a\\ b \end{bmatrix} $