How do I intuitively understand what this linear transformation matrix is? $\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}$
I know how to get the product when given another matrix. But how do I know what this matrix is doing simply by looking at it?
 A: Note that 
\begin{equation}
\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} y \\ -x \end{bmatrix}.
\end{equation}
You can draw a picture to see that the vector $\begin{bmatrix} y \\ -x \end{bmatrix}$ is what you get when you rotate $\begin{bmatrix} x \\ y \end{bmatrix}$ clockwise by 90 degrees.
You could also note that
\begin{align}
\left\langle \begin{bmatrix} x \\ y \end{bmatrix}, \begin{bmatrix} y \\ -x \end{bmatrix} \right \rangle &= xy - yx \\
&= 0
\end{align}
and
\begin{align}
\left\| \begin{bmatrix} y \\ -x \end{bmatrix} \right \| &= \sqrt{y^2 + x^2} \\
&= \left\|\begin{bmatrix} x \\ y \end{bmatrix} \right\|.
\end{align}
This shows that $\begin{bmatrix} y \\ -x \end{bmatrix}$ is a rotation of $\begin{bmatrix} x \\ y \end{bmatrix}$ by 90 degrees.
A: Just look how it works on the canonical base $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$.
The results
$\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \end{bmatrix} $ 
$\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $ 
are just one column selected from the matrix. These can be seen as a base in new vector space. Draw both bases, and you'll see how the original space was modified. 
In this example: rotation by 90 degrees clockwise. 
In general every linear operation just need to be known on base vectors. Look what it does with the canonical base, and you'll see everything there is to see.
