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  1. (6 points) Let $V$ be the vector space of polynomials of degree at most $99$ with real coefficients. Define a linear map $$T:V\to\mathbb{R}^{1000},\quad T(p)=(p(1),p(2),\ldots,p(1000)).$$ a) Find the dimension of the null space of $T$.

For this problem, I try to write $p(x) \in V$ as $p(x) = z_{1000}(x) \times q(x)$, where $$z_{1000}(x)=(x-1)(x-2)\cdots(x-1000),$$ and $$q(x)= a_{-901}x^{-901} + a_{-902}x^{-902} +\cdots+ a_{-1000}x^{-1000},\quad a_i \in R.$$

This shows that the basis of $\operatorname{null}(T)$ consists of $100$ polynomials $$(z_{1000}(x)\times x^{-901}, z_{1000}(x)\times x^{-902},\ldots,z_{1000}(x)\times x^{-1000}).$$

Thus $\operatorname{null}(T)$ has dimension $100$. This is obviously wrong since then $\operatorname{null}(T)=V$, but there are many counter-examples to this.

Where did I go wrong?

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3 Answers 3

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Note that $$T(p)=0\iff p(1)=...=p(1000)=0\iff p(x)=0$$

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Your polynomials (such as $z_{1000}(x)\times x^{-901}$) are no polynomials at all.

If $p(x)\in\ker T$ then $p(x)$ has $1\,000$ roots and degree at most $99$. Therefore, $p(x)=0$ and so $\ker T=\{0\}$.

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  • $\begingroup$ Could you please explain why expressions such as $z_{1000}(x)×x^{−901}$ are not polynomials? $\endgroup$
    – ensbana
    Apr 14, 2018 at 17:47
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    $\begingroup$ @ensbana Beacuse polynomials are sums of expressions of the type $ax^k$, with $k\in\mathbb{Z}^+$. $\endgroup$ Apr 14, 2018 at 17:48
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If you had such a polynomial say $p \in V$, then means that $p(1), ..., p(1000)$ are all zero. Thus a polynomial of degree at most $99$ has at least $1000$ roots... Impossible, unless $p=0$; so ${\rm ker}(T)=\{0\}$ and its dimension is zero.

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