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Problem 4 from https://math.berkeley.edu/~ogus/Math_54-07/Exams/midsol1.pdf

$\beta$ 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 $R^2$

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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 $x^2-1$.

(b) and (c): Just substitute the given basis elements (in the given order) into the map, and find the coordinates of the result w.r.t. the given target basis, and write these into the columns.

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 $\pmatrix{4\\-1}$.

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  • $\begingroup$ why does $p(-1) = (-1)^2$? $p(x) =d+cx+bx^2+ax^3$. So $p(-1) = d-c +b -a.$ $\endgroup$ – larry Apr 13 '15 at 20:56
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    $\begingroup$ You substitute $p=x^2$ in the place of $p$, so evaluate $p(-1)$, and while evaluating you substitute $x=-1$ into $p=x^2$. $\endgroup$ – Berci Apr 13 '15 at 21:02

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