Find matrix of $T(z) = (4 + 9i)z$ for basis $\{2 + 2i, 2 + 4i\}$ How can we find the matrix $A$ of the linear transformation $T(z) = (4 + 9i)z$ from $\mathbb C$ to $\mathbb C$ with respect to the basis $\{2 + 2i, 2 + 4i\}$?
 A: Let v_1 and v_2 be your basis vectors then row reduce the left and you answer is on the right:
$$\begin{bmatrix} v_1 \ v_2 \ | \ T(v_1) \ T(v_2) \end{bmatrix}$$
A: $T(2+2i) = (4 + 9i)(2 + 2i) = -10 + 26i$
$T(2+4i) = (4 + 9i)(2 + 4i) = -28 + 34i$
So you now know $TB^{-1}$.  Compute $BTB^{-1}$ to find $[T]_B$ where $B^{-1}$ is the basis in a matrix as columns, the inverse matrix to the change of basis matrix $B$.
A: Let's say you are building your 2-by-2 matrix $M$ so as to multiply with 1-by-2 row vectors of real numbers.
In this setup, your first basis vector has coordinates $[1,0]$ and the second $[0,1]$.
When you multiply $[1,0]M$ and $[0,1]M$ you should get the coordinates of their images, respectively.
Getting the value of their images is not hard: compute $(2+2i)(4+9i)$ and $(2+4i)(4+9i)$.
The next step would be to write these two images as linear combinations of your two basis vectors, and then arrange the coefficients in $M$ correctly to model the transformation.
I purposefully omitted the details of the last part because these are steps you really need to carry out yourself so that you can get the hang of how transformations are expressed as matrices. I think the rest of what I wrote gives a complete roadmap!
