Linear transformations, matrices and change of basis My question is in regards to matrices and basis. For example,the matrix $B$
$$ \left[
\begin {matrix} 
 1 & 2 \\ 
 1 & 0 \\
\end {matrix}
\right]
$$
and the basis $$w_1 = [1,0]$$$$w_2 = [0,1]$$
Given $B$ is the matrix representation of a linear transform $g$, my question is in regards to what the matrix represents. Is the matrix a particular combination of basis vectors, and why does $g(w_1) = [1,1]$ 
How can I think of this matrix $B$ in relation to the rotation matrix 
$$ \left[
\begin {matrix} 
 \cos \theta & \sin \theta\\ 
 -\sin \theta & \cos \theta\\
\end {matrix}
\right]
$$
of which I'm familiar with?
Thanks for the help
 A: While a rotation is a type of linear transformation, not all linear transformations are rotations. The matrix given is not a rotation. However, there is a fairly nice way to picture the image of a $2\times 2$ matrix as a linear transformation. The way to do it is to look at the matrix
$$
\begin{bmatrix}
a & c \\
b & d
\end{bmatrix}
.$$
This linear transformation will send the standard unit vector $\begin{bmatrix} 1 \\ 0\end{bmatrix}$ to the position vector  represented by $\begin{bmatrix} a \\ b\end{bmatrix}$. Similarly, under this transformation, the standard unit vector $\begin{bmatrix} 0 \\ 1\end{bmatrix}$ will go to the position vector  represented by $\begin{bmatrix} c \\ d\end{bmatrix}$. In other words, the first column is where the "$x$" unit vector lands after the transformation, and the second column is where the "$y$" unit vector lands. Using this approach, you might be able to interpret the matrix you gave as a flip about the line $y=x$, followed by a horizontal shear one unit to the right. Indeed
$$
\begin{bmatrix}
2 & 1 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
=
\begin{bmatrix}
1 & 2 \\
1 & 0
\end{bmatrix}
,$$
which is a correct interpretation of this statement.
A: Applying the matrix to the column vector $\begin{pmatrix} a \\ b \\ \end{pmatrix}$ you get 
$\begin{pmatrix} 1 & 2 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a \\ b \\ \end{pmatrix}=\begin{pmatrix} a+2b \\ a \end{pmatrix} $
So $g$ which is the linear transformation associated with $B$  is 
$g\begin{pmatrix} a \\ b \\ \end{pmatrix}=\begin{pmatrix} a+2b \\ a \end{pmatrix}$    
So technically $B$ represents this linear transformation with respect to the standard basis vectors $w_1$ and $w_2$. You can now easily see that $g(w_1)=(1,1)$
Finally if you want to know how $w_1$ and $w_2$ are related to $B$, just apply $g$ to $w_1$ and $w_2$ and you will realize that the columns of $B$ are exactly $g(w_1)$ and $g(w_2)$
A: Writing vectors as columns would be more convenient.
$T$ is a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$; $e_1, e_2,...,e_n$ is the basis of $\mathbb{R}^n$; $b_1,b_2,...,b_m$ is the basis of $\mathbb{R}^m$; $T(x)$ is the coordinates of the image of $x \in \mathbb{R}^n$ under the basis $b_1,...,b_m$.
Then, the matrix of $T$ is
$$
M(T) = \begin{bmatrix} T(e_1)&T(e_2)& ... &T(e_n) \end{bmatrix}.
$$
For example, $T$ is the rotation on $\mathbb{R}^2$. We take the basis $\{\begin{bmatrix} 1\\ 0\end{bmatrix},\begin{bmatrix} 0\\ 1\end{bmatrix}\}$.
$$
T(\begin{bmatrix} 1\\ 0\end{bmatrix}) = \begin{bmatrix} cos\theta\\ sin\theta\end{bmatrix},
$$
$$
T(\begin{bmatrix} 0\\ 1\end{bmatrix}) = \begin{bmatrix} -sin\theta\\ cos\theta\end{bmatrix}.
$$
So the matrix is
$$
\begin{bmatrix}
cos\theta & -sin\theta \\
sin\theta & cos\theta
\end{bmatrix}.
$$
If we write vectors as rows, we could just take the transposition. In this condition, the matrix is what you mentioned in the question.
$$
\begin{bmatrix}
cos\theta & sin\theta \\
-sin\theta & cos\theta
\end{bmatrix}.
$$
