# Find basis for kernel and matrix representation

$\beta$ is a basis of $P_3$, the set of all polynomials of at most degree 3.$\beta = (x^0,x^1,x^2,x^3)$. Let $T$ be a linear transformation from $P_3$ to $R^2 = (p(-1),p(1))^T$.

(a) Find a basis for the kernel of $T$. (I know the answer, just not sure how it was gotten.)

(b) Find the matrix $T'$ basis $\beta$ and the standard basis for $R^2$.

A secondary question: (c) How do you find the matrix $T'$ basis $\beta = (x^3,x^2,x^1,x^0)$ and the standard basis (1,2)^T and (3,7) for $R^2$

So, $T$ is the linear map $p\mapsto \pmatrix{p(-1)\\p(1)}$.
(a) First find the kernel: this is the set of those polynomials $p$ which satisfy $p(-1)=0$ and $p(1)=0$ simultaneously. I.e., $p$ contains a linear factor $x+1$ and also an $x-1$, hence $p$ is a multiple of $x^2-1$.
E.g. $T(x^2)=\pmatrix{(-1)^2\\1^2}=\pmatrix{1\\1}=1\cdot\pmatrix{1\\0}+1\cdot\pmatrix{0\\1}=4\cdot\pmatrix{1\\2}-1\cdot\pmatrix{3\\7}$,
so the third column of the requested matrix in (b) is $\pmatrix{1\\1}$ and the second column of that in (c) is $\pmatrix{4\\-1}$.
• why does $p(-1) = (-1)^2$? $p(x) =d+cx+bx^2+ax^3$. So $p(-1) = d-c +b -a.$ – larry Apr 13 '15 at 20:56
• You substitute $p=x^2$ in the place of $p$, so evaluate $p(-1)$, and while evaluating you substitute $x=-1$ into $p=x^2$. – Berci Apr 13 '15 at 21:02