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In my textbook there is an example of how the kernel can be found, but I am very unsure how the argument for finding the kernel holds

Let $n\gt1$ describe a natural number and let $\alpha \in \mathbb{R}$ be a real scalar. Observe the subset $$ V = \{p\in P_n(\mathbb{R}):p(\alpha)=0\}$$ of the real vector space $P_n(\mathbb{R})$

Now observe the map $$T: P_{n-1}(\mathbb{R}) \to P_n(\mathbb{R})$$ $$p \to p\cdot(X-\alpha)$$

Show that the $\operatorname{Ker}(T)$ is the zero-vector space and show that the image of $T$ has the dimension $n-1$

The $\operatorname{Ker}(T)$ is given by: $$\operatorname{Ker}(T) = \{p\in p_{n-1}(\mathbb{R})\mid p\cdot(X-\alpha)=0\}$$

It must then be shown that $p\cdot(X-\alpha)=0$, which is only the case when $p=0$, which proves that the $\operatorname{Ker}(T)=\{0\}$.

In order to find show that the dimension $\operatorname{Im}(T)=n-1$ it is possible to use the rank-nullity theorem as $\operatorname{Ker}(T)=\{0\}$: $$\dim(\operatorname{Im}(T)) + \dim(\operatorname{Ker}(T)) = \dim(P_{n-1}(\mathbb{R})$$ Therefore, $\dim(\operatorname{Im}(T)) = \dim(P_{n-1}(\mathbb{R})=n-1$

What I don't really understand is the argument for proving that $\operatorname{Ker}(T)=\{0\}$. Why is it only possible when $p=0$ and not when $(X-\alpha)=0$? I hope that someone can explain the part about how the kernel is found in greater detail.

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  • $\begingroup$ $P_n(\mathbb{R})$ is the set of polynomials of degree $n$ with coefficients in $\mathbb{R}$? $\endgroup$ Commented May 21, 2018 at 9:56
  • $\begingroup$ That is because $X-\alpha$ is not $0$. You're confusing polynomial functions and (true) polynomials. Here $X$ is an indeterminate. $\endgroup$ Commented May 21, 2018 at 9:58
  • $\begingroup$ You're looking for polynomials $p$ s.t. $T(p)=p(X-\alpha)=0,$ not for values $X.$ $\endgroup$ Commented May 21, 2018 at 10:05
  • $\begingroup$ What textbook are you using? $\endgroup$ Commented May 21, 2018 at 10:22
  • $\begingroup$ Yes $P_n(\mathbb{R})$ is a set of polynomials of degree $n$ with coefficients in $\mathbb{R}$. So because we have to find the zero polynomial it must be the case that $p=0$? (The textbook we are using was made by our professor) $\endgroup$ Commented May 21, 2018 at 10:29

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Note that as a zero polynomial $$p\cdot(X-\alpha)=0$$ if and only if $p(x) =0$ because $x-\alpha $ is a polynomial of degree $1$ and it is not a zero polynomial.

Note that a zero polynomial $p(x)$ is a polynomial which is satisfies $p(x)=0$ for all values of $x$.

$p(x)=x-\alpha$ is zero only at $x=\alpha $

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