Looking for a conceptual understanding of a rotation matrix transformation I'm new to Linear Algebra and I'm having a hard time wrapping my head around linear transformations, specifically a rotation.
From Anton's book (Elementary Linear Algebra, 11th Edition) he states:
$T(e_1) = T(1,0) = (\cos\theta, \sin\theta)$
and
$T(e_2) = T(0,1) = (-\sin\theta, \cos\theta)$
and
$A = [T(e_1) | T(e_2)] = \begin{bmatrix} 
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$
When I rotate a vector $\begin{bmatrix} x\\y \end{bmatrix}$ I get 
$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} x \cdot\cos\theta \, - \, y \cdot\sin\theta \\
x \cdot\sin\theta \, + \, y \cdot\cos\theta
\end{bmatrix}$
Correct me if I'm wrong, but I thought that column 1 of $A$ $\begin{bmatrix} \cos\theta \\
\sin\theta
\end{bmatrix}$, holds the 'x' values and column 2 holds the 'y' values.  What I'm confused about is why does $x'$ contain both an $x$ component and a $y$ component?
When I transform $\begin{bmatrix}1 \\ 0 \end{bmatrix}$ by a rotation $\theta$, the new $x$ value is just $\cos\theta$ while the new $y$ value is just $\sin\theta$.  I don't understand why with the matrix transformation, $x'$ and $y'$ get both $x$ and $y$ components summed together.
I hope I'm making sense.
 A: Your typical point with polar coordinates $r$ and $\phi$ and has vector
$$\begin{bmatrix}
r\cos\phi\\r\sin\phi
\end{bmatrix}.$$
Multiplying $A$ into this gives
$$\begin{bmatrix}
r\cos\theta\cos\phi-r\sin\theta\sin\phi\\r\sin\theta\cos\phi+r\cos\theta\sin\phi
\end{bmatrix}
=\begin{bmatrix}
r\cos(\theta+\phi)\\r\sin(\theta+\phi)
\end{bmatrix}.$$
It's at the same distance from the origin, but rotated anticlockwise by an angle $\theta$.
A: Any vector in $\Bbb{R}^2$ can be represented as follows
$$v=\left(\begin{array}{c}
x\\
y
\end{array}\right)=|| v|| \left(\begin{array}{c}
\cos(\phi)\\
\sin(\phi)
\end{array}\right).$$
If $$T=\left(\begin{array}{cc}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array}\right)
$$
Then $$Tv= || v||\left(\begin{array}{cc}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{array}\right) 
\left(\begin{array}{c}
\cos(\phi)\\
\sin(\phi)
\end{array}\right)=||v||\left(\begin{array}{c}
\cos(\phi)\cos(\theta)-\sin(\phi)\sin(\theta)\\
\cos(\phi)\sin(\theta)+\sin(\phi)\cos(\theta)
\end{array}\right)$$
Now, you will recognize this as the angle sum formula. That is
$$\left(\begin{array}{c}
\cos(\phi)\cos(\theta)-\sin(\phi)\sin(\theta)\\
\cos(\phi)\sin(\theta)+\sin(\phi)\cos(\theta)
\end{array}\right)=\left(\begin{array}{c}
\cos(\phi+\theta)\\
\sin(\phi+\theta)
\end{array}
\right)
$$ 
so ultimately you can see that this matrix is simply rotating by adding the appropriate angle, and this is why you see the original components of the vector after applying the linear transformation.
A: 
why does $x′$ contain both an $x$ component and a $y$ component?

This is true for any matrix transformation. If the transformation is represented by matrix
$$
A:=\left[\begin{matrix}a&b\\
c&d\\
\end{matrix}\right]
$$
then for an arbitrary vector $\begin{bmatrix}x \\ y \end{bmatrix}$ you get
$$
\begin{align}
\begin{bmatrix}x' \\ y' \end{bmatrix}&=T\begin{bmatrix}x \\ y \end{bmatrix}\\
&=T\left(x\begin{bmatrix}1 \\ 0 \end{bmatrix} + y\begin{bmatrix}0 \\ 1 \end{bmatrix}\right)\\
&=xT\left(\begin{bmatrix}1 \\ 0 \end{bmatrix}\right) + yT\left(\begin{bmatrix}0 \\ 1 \end{bmatrix}\right)\\
&=x\begin{bmatrix}a \\ c \end{bmatrix} + y\begin{bmatrix}b \\ d\end{bmatrix}\\
&=\begin{bmatrix}ax \\ cx \end{bmatrix} + \begin{bmatrix}by \\ dy\end{bmatrix}\\
&=\begin{bmatrix}ax+by \\ cx+dy \end{bmatrix}.
\end{align}
$$
When $\begin{bmatrix}x \\ y \end{bmatrix}$ is a unit vector, say $\begin{bmatrix}x \\ y \end{bmatrix}=\begin{bmatrix}1 \\ 0 \end{bmatrix}$, all but one of the coordinates is zero so the effect is to pick out just one column of $A$.   Otherwise more than one column of $A$ comes into play.
A: Column 1 contains the $x$ and $y$ values for the vector representing where the transformation takes $e_1$. For example, if the transformation is counterclockwise rotation by $\pi/4$ ($45$ degrees) that vector goes to 
$$
(\sqrt{2}/2,\sqrt{2}/2) = (\cos(\pi/4), \sin(\pi/4)).
$$
It's the first row, not the first column, that contains the $x$ values. Thinking about the rows is generally less informative than thinking about the columns.
I think your problem is not conceptual, it's with the notation. You seem to understand what's going on.
A: I found the answer.  I kept thinking about this and drew out the angle rotation using a ruler, compass, and protractor to see if I could find why the rotation formula is the way it is.  I don't have the time to learn TikZ but basically this is where I got to:

We are originally at angle $\alpha$ and want to rotate $\beta$ degrees.  In other words, we want to get from $\overline{AE}$ to $\overline{AF}$.
The $x$ coordinate of $\overline{AE} = \overline{AC}$. The $y$ coordinate of $\overline{AE} = \overline{CE}$.
After a rotation of $\beta$ degrees, the new $x = x' = \overline{AB}$ and the new $y = y' = \overline{FB}$.

$x' = \overline{AB} = \overline{AC} - \overline{BC}$
$\overline{BC} = \overline{DE}$
$\triangle ACE: \cos\alpha = \frac{\overline{AC}}{\cos\beta}$
$\overline{AC} = \cos\alpha\cos\beta$
$\angle DFE = \angle CAE$
$\triangle DEF: \sin\alpha = \frac{\overline{DE}}{\sin\beta}$
$\overline{DE} = \sin\alpha\sin\beta$
$x' = \cos\alpha\cos\beta - \sin\alpha\sin\beta$

$y' = \overline{FB} = \overline{FD} + \overline{DB}$
$\overline{DB} = \overline{CE}$
$\triangle DEF: \cos\alpha = \frac{\overline{FD}}{\sin\beta}$
$\overline{FD} = \cos\alpha\sin\beta$
$\triangle ACE: \sin\alpha = \frac{\overline{CE}}{\cos\beta}$
$\overline{CE} = \sin\alpha\cos\beta$
$y' = \cos\alpha\sin\beta + \sin\alpha\cos\beta$
