Hard change of basis/ linear transformation problem

Question

Let $T : \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation, such that $T (2,1)= (0,0)$ and $T (-1,2)= (1,3)$. What is the standard matrix A of T?

I understand that you need to convert the given basis to the standard basis, however do not really get how to do that. I also am sort of confused why the basis is $(2,1)$ and $(1,3)$ (my friend told me to use that basis, however I still don't really understand why).

I have looked all over the internet for a problem like this, but I am having trouble. I would really appreciate any help, thanks so much!!!

Answer 1

Let $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$

We are told that $A\cdot \begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ and that $A\cdot\begin{bmatrix}-1\\2\end{bmatrix}=\begin{bmatrix}1\\3\end{bmatrix}$

Reworded, this implies the following system of equations:

$\left\{\begin{array}{rl} 2a+b&=0\\2c+d&=0\\-a+2b&=1\\-c+2d&=3\end{array}\right\}$

Now... you should have learned techniques on how to solve that system, for example with Gaussian Elimination.

Answer 2

"I understand that you need to convert the given basis to the standard basis, however do not really get how to do that."

Simply: $(2,1)=2\cdot(1,0)+1\cdot(0,1)$, $(1,3)=1\cdot(1,0)+3\cdot(0,1)$.

"I also am sort of confused why the basis is (2,1) and (1,3) (my friend told me to use that basis, however I still don't really understand why)."

I think You should ask Your friend.

For a matrix in a standard basis You have to represent $T(1,0)$ and $T(0,1)$ as linear combinations of $(1,0)$ and $(0,1)$ and then to write the coefficients of these combinations as columns. So you have to calculate $T(1,0)$ and $T(0,1)$. This goes so: $T(1,0)=0.2T(5,0)=0.2(T(4,2)-T(-1,2))=0.4T(2,1)-0.2T(-1,2)=$

$=(0,0)-0.2(1,3)=(-0.2,-0.6)$.

So You have Your first column. The second column goes in a similar way.