# 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

• I wrote a little article with examples of shuffling vectors in a base. It is not directly answering your specific question but I think it can help you figure out what the columns of a matrix represents. math.stackexchange.com/questions/2259746/…
– zwim
Dec 19 '17 at 3:47

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.

• Thanks. Is it fair to say that matrices, being linear transformations can be expressed in equation form? Do the columns express combinations of the standard basis? Would that mean the above matrix $B$ would have two equations, one for the transformed $x$ (in terms of the standard basis), which is $x$ and $y$ and one for the transformed $y$ would be twice the standard $x$ basis? Dec 19 '17 at 7:29

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 $g(w_2)$

• Okay, so through mere matrix multiplication of the vector $w1$ and $g$ I get $[1, 1]$? Dec 19 '17 at 7:19
• @SuperJumbo No . Since $w_1=(1,0)$, plug in $a=1$ and $b=0$ into the formula for $g$. That's how we evaluate linear transformations. It's like evaluating regular functions Dec 19 '17 at 7:27
• Okay, thanks : ) Dec 20 '17 at 1:05

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.

$$\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}.$$