For this question I need to find what matrix A is, but I don't know what to do it since the question only has null space and span. I thought A was [-1 -1] but I'm sure it's wrong.
Could somebody help me with that? Thank you so much!
I want to display the image here but the website it needs 10 reputations to do so. Sorry for the inconvinence :(
Seen $A$ as linear transformation, you have that $A(1,-2)=(-1,2)$ and $A(1,1)=0$. Note that this is enough to determine completely our transformation $A$ since ${(1,-2),(1,1)}$ is a basis of $\mathbb{R}^2
Let's compute what is $A$ explicitely, i.e. let's compute $A(x,y)$ as a formula of $x,y$. To do this, we express $(x,y)$ as a linear combination of $(1,-2)$ and $(1,1)$, i.e. we need to find $a,b$ (in terms of $x,y$) such that $$(x,y)=a(1,-2)+b(1,1)$$ This leads to two (linear) equations \begin{align*} x=a+b\\ y=-2a+b \end{align*} Solving this equation, you get $a=\frac{x-y}{3}$ and $b=\frac{2x+y}{3}$.
Now, we can compute $A(x,y)$ as follows:
Hence \begin{align*}A(x,y)&=A(a(1,-2)+b(1,1))\\ &=aA(1,-2)+bA(1,1)\\ &=\frac{x-y}{3}(-1,2)+\frac{2x+y}{3}(0,0)\\ &=\left(-\frac{1}{3}x+\frac{1}{3}y\,,\, \frac{2}{3}x-\frac{2}{3}y\right). \end{align*}
So, as a matrix $$A=\left(\begin{array}{cc}-\frac{1}{3}&\frac{1}{3}\\ \frac{2}{3}&-\frac{2}{3} \end{array}\right)$$ because $$A{x\choose y}=\left(\begin{array}{cc}-\frac{1}{3}&\frac{1}{3}\\ \frac{2}{3}&-\frac{2}{3} \end{array}\right){x\choose y}={-\frac{1}{3}x+\frac{1}{3}y\choose \frac{2}{3}x-\frac{2}{3}y}$$