Let $P= \mathbb{C}[x] $ be the vector space of polynomials with complex coefficents. We define the following linear recurrence $p_{0}(x)=1$, $p_{1}(x)=x$, $p_{n}(x)=p_{n-1}(x)-p_{n-2}(x)$ for $n \geq 2$
How do I formulate this in the form $y_{n}=Ay_{n-1}$?
It is clear from the definition of $\bigl(p_n(x)\bigr)_{n\in\mathbb{Z}_+}$ that$$(\forall n\in\mathbb{Z}_+):\deg p_n(x)\leqslant1.$$So, let us work in the space $\mathbb{C}_1[x]$ of all polynomials $p(x)\in\mathbb{C}[x]$ such that $\deg p(x)\leqslant1$. If such a linear map $A$ existes, then $A(1)=x$ and $A(x)=-1+x$. But, since $\{1,x\}$ is a basis of $\mathbb{C}_1[x]$, there is one and only one such map: $A(a+bx)=ax+b(-1+x)=-b+(a+b)x$. Does it work? You can see that $A(-1+x)=-1$. Also, $A(-1)=-x$, $A(-x)=1$, and $A(1)=x$ (by definition), and we're back to the first polynomial. And you can easily see that your sequence is also$$1,x,-1+x,-1,-x,1,\ldots$$