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Let $\mathbb{R}_3[x]$ be a linear vector space of polynomials of degree $\leq 3$ and let $$f: \mathbb{R}_3[x] \ni p \mapsto (p(0), p(1), p(2)) \in \mathbb{R}^3.$$ Find the dimensions of kernel and image of $f$ and any basis of the kernel.

First, I examine the kernel space.

Let $p(x) = ax^3 + bx^3 + cx + d \in \mathbb{R}_3[x]$ of any real $a, b, c, d$. Then $$p \in \ker(f) \iff f(p) = (0, 0, 0)$$ and thus I obtain the system of equations:

$$\begin{cases} p(0) = d = 0 \\ p(1) = a + b + c = 0 \Rightarrow c = - a - b \\ p(2) = 8a + 4b + 2c = 0 \end{cases}$$ Substituting $c$ to the last equation: $$8a + 4b - 2a - 2b = 0 \Rightarrow b = -3a$$ Hence $c = 2a$ and

$$p \in \ker(f) \iff p(x) = ax^3 - 3ax^2 + 2a = a(x^3 - 3x^2 +2)$$

Therefore, since $a$ is any real number, I suspect that $$\ker(f) = \text{span}\left(\{x^3 - 3x^2 + 2\}\right)$$

Is that correct? Would this follow that $\{x^3 - 3x^2 + 2\}$ is the kernel basis, leading to $\dim(\ker(f)) = 1$?

I would then conclude that

$$\dim(\text{im}(f)) = \dim\left(\mathbb{R}_3[x]\right) - \dim(\ker(f)) = 4 - 1 = 3$$

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2 Answers 2

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It is almost correct, but the conclusion should be that $\ker f=\langle x^3-3x^2+2x\rangle$.

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  • $\begingroup$ Oh, right, sorry, that was a typo. Thanks for noticing. $\endgroup$
    – Angie
    Jul 4, 2018 at 18:23
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$\ker f=\operatorname{span}\{x(x-1)(x-2)\}$ has dimension $1$. Thus, by the Rank-nullity theorem, the image of $f$ has dimension $3$. So you are correct (other than losing an $x$ in $x^3-3x^2+2\color{red}x$...)

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