Rotation of vector I'm reading Anton's Elementary Linear Algebra.  I have come upon the rotation matrix.
$\begin{bmatrix} \cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \end{bmatrix}$
They start the discussion with the fact that $T(e_1) = T(1,0) = (\cos \theta, \sin \theta)$ and $T(e_2) = T(0,1) = (-\sin \theta, \cos \theta)$.
This makes sense to me.  But why do you need both $e_1$ and $e_2$ ?   What about $e_3$ to rotate a vector in 3 dimensions?
EDIT
More specifically, I'm wondering about rotation in 3 dimensions.
About the z-axis:
$\begin{bmatrix} \cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1\end{bmatrix}$
About the x-axis:
$\begin{bmatrix} 1 & 0 & 0\\
0 & \cos \theta & -\sin \theta \\
0 & \sin \theta & \cos \theta\end{bmatrix}$
About the y-axis:
$\begin{bmatrix} \cos \theta & 0 & \sin \theta \\
0 & 1 & 0 \\
-\sin \theta & 0 & \cos \theta\end{bmatrix}$
I see that both z-axis rotation and x-axis rotation follow the pattern of $\begin{bmatrix} \cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \end{bmatrix}$
Why is the rotation about the y-axis different?
 A: The first part of your question makes no sense to me.  However, with your edit:

Why is the rotation about the y-axis different?

The things to notice is how exactly the $2D$ rotation relates to the $3D$ rotation. With the rotation about the $z$-axis, $e_1,e_2$ play the role of the "$x,y$ axes" from $2D$.  If we switched $e_1,e_2$, that is, if instead we used $e_2$ as our "$x$-axis" and $e_1$ as our "$y$-axis", we would end up with
$$
\pmatrix{\cos \theta & \sin \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0 & 0  & 1}
$$
With the rotation about the $x$-axis, $e_2,e_3$ play the role of the $x,y$ axes in $2D$.  Because we keep them in numerical order, we get the expected submatrix.
With the rotation about the $y$-axis, $e_3$ plays the role of the $x$-axis and $e_1$ plays the role of the $y$-axis.  That is, the numerically higher $e_3$ comes first.  If we switched these roles, we would have ended up with the matrix
$$
\pmatrix{\cos \theta & 0 & - \sin \theta\\ 0 & 1 & 0\\ \sin \theta & 0 & \cos \theta}
$$
which is what you were expecting.
So why is $e_3,e_1$ the "correct order"?  It has to do with what "counterclockwise" means in $3$ dimensions.  $e_1,e_2,e_3$ form a right-handed coordinate system, so a counterclockwise rotation in the $1,2$ plane should go from $e_1$ to $e_2$, and similarly a counterclockwise rotation in the $1,3$ plane should go from $e_3$ to $e_1$.
A: The easiest way to understand what a linear transformation does is to pay attention to what it does to a fixed basis. 
In this case, they've just chosen the most familiar basis $e_1$ and $e_2$. By drawing where these two things land and comparing that to where they started, you can see what the transformation did to the plane.

What about $e_3$ to rotate a vector in $3$ dimensions?

Well, yes, if we were talking about a $3\times 3$ rotation matrix, then you could map all of $e_1,e_2,e_3$ to their images and do the same thing. But that is not relevant for a rotation of the plane...
A: I think I figured it out.  I think this has to do with the fact that this is from the computer graphics section of the book and in this world the z-axis is defined as negative z pointing into the screen, and positive z pointing away from the screen.
During a rotation about the z-axis, $e_1$ is rotated into the 1st quadrant as $\theta$ goes from $0^\circ$ in the positive direction (counterclockwise).  $e_2$ is rotated into the second quadrant in the same manner. That is why $T(e_1) = T(1,0) = (\cos \theta, \sin \theta)$ and $T(e_2) = T(0,1) = (-\sin \theta, \cos \theta)$.
However, a rotation about the y-axis in computer graphics rotates $e_1$ towards the -z direction (into the 4th quadrant) or from left to right when looking at an object from the conventional viewer's perspective (on the positive z axis looking towards the negative z axis).  And so $T(e_1) = T(1,0) = (\cos \theta, -\sin \theta)$ and $T(e_2) = T(0,1) = (\sin \theta, \cos \theta)$.
