How to give a geometrical interpretation of a linear transformation? The question reads as follows:
Let
$$A=\begin{bmatrix}-1/2 & -\sqrt{3}/2\\\sqrt{3}/2& -1/2\end{bmatrix}$$
and let $T = \Bbb{R}^2 \to \Bbb{R}^2$ be the linear transformation described as $T(x) = Ax$

Give a geometrical interpretation of the linear transformation and calculate A$^{2021}$

The solution says that $A$ rotates the vectors in an angle of $2\pi/3$, but how do you see that? And what does the matrix even mean? Does it describe the evolution of the first vector column as it rotates?
 A: Every matrix of the form
$$\begin{pmatrix}a & -b \\b & a \\\end{pmatrix}
\in \Bbb R^{2\times2}\tag 1$$
can be regarded as complex number $z=a+ib\in\Bbb C$ by identifying
$$
1\leftrightarrow\begin{pmatrix}1 & 0 \\0 & 1 \\\end{pmatrix}
\qquad\text{and}\qquad
i\leftrightarrow\begin{pmatrix}0 & -1 \\1 & 0 \\\end{pmatrix}
$$
Notice that $i^2 = -1$.
As any multiplication with a complex number is a zoom-and-rotate, this is also the case for your choice of $z=A= (-1 +\sqrt 3)/2$.  Notice that $|A|=\sqrt{a^2+b^2}=1$, thus a multiplication by $A$ is a rotation with no zoom. Also notice that $A^3=1$ because it is a 3rd root of unity and therefore
$$A^{2021} = A^{2021\text{ mod }3}= A^2 = A^{-1} = A^{T}
$$
The geometric interpretation is an (anti-)clockwise rotation by 120°, where the handed-ness of your unit vectors determines the handed-ness of the rotation.  With usual depictions of compex numbers, the rotation is 120° anti-clockwise.

If you don't want complex numbers in that context, then notice that any matrix of the form $(1)$ can be written as
$$\begin{pmatrix}a & -b \\b & a \\\end{pmatrix}
= \sqrt{a^2+b^2} \begin{pmatrix}\cos\phi & -\sin\phi \\\sin\phi & \cos\phi \\\end{pmatrix}
$$
where again $\sqrt{a^2+b^2}=1$ and $\phi = 2\pi/3\text{ mod }2\pi$. Thus, again we see that $A$ is a rotation by 120° (mod 360°).
A: Here I'm gonna use $V$ to denote the vector you called $x$.
Pick any vector $V=\begin{bmatrix} x \\ y \end{bmatrix}$.
Compute the angle between $V$ and $A\cdot V$.
You'll find out this angle is $\frac{2\pi}{3}$. This is what the matrix $A$ does when you multiply it by a vector on the right: it rotates it clock-wise by an angle of $\frac{2\pi}{3}$.
This matrix is a particular case of a  rotation matrix on the plane.
A: The image of the vector $\left(\begin{matrix}1\\0\end{matrix}\right)$ is the first column of your matrix.
The image of the vector $\left(\begin{matrix}0\\1\end{matrix}\right)$ is the second column of your matrix.
As is the case with any 2x2 matrix, you can get an idea of what the transformation does simply by sketching these two vectors comprising the first and second columns of the matrix.
To check you can also compare the matrix with the standard form of a rotation anticlockwise by angle $\theta$ which is $$\left(\begin{matrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{matrix}\right)$$
Now it should be clear that is is a rotation anticlockwise by $120^o$.
I assume you can do the last part.
