By given the following linear transformation: $T(p(x))=p'(x)$,$T:P_3[\mathbb{R}]\rightarrow P_3[\mathbb{R}]$, find a basis and the dimension of the kernel.

$Solution.$ \begin{align*} \ker T & =\left\{p( x) \in P_{3}[\mathbb{R}]\Bigl| p'( x) =0\right\} \\ &=\left\{ax^{3} +bx^{2} +cx+d\Bigl| 3ax^{2} +2bx+c=0,a,b,c,d\in \mathbb{R}\right\}\\ & =\left\{ax^{3} +bx^{2} +cx+d\Bigl| c=-3ax^{2} -2bx,a,b,c,d\in \mathbb{R}\right\}\\ & =\left\{ax^{3} +bx^{2} +\left( -3ax^{2} -2bx\right) x+d\Bigl| a,b,d\in \mathbb{R}\right\}\\ & =\left\{a\left( x^{3} -2x^{2}\right) +b\left( x^{2} -2x\right) +d\Bigl| a,b,d\in \mathbb{R}\right\}\\ & =\operatorname{Span}\left\{x^{3} -2x^{2} ,x^{2} -2x,1\right\}\end{align*} in addition, the isomorphic vectors to the polynomials are linearly independent, can be easily checked and easy to see, so they are a basis for the kernel. Thus, $$B_{\ker T} =\left\{x^{3} -2x^{2} ,x^{2} -2x,1\right\} \Longrightarrow \dim\ker T=3$$

However, it is clear that this isn't true since the only polynomial who gives $p'(x)=0$ is $p(x)=d$, so the basis is actually: $$B_{\ker T} =\left\{1\right\} \Longrightarrow \dim\ker T=1$$

and it's easy to see.

Perhaps I'm finding here a basis for the $x$'s? because I can't find why I get the wrong polynomials in the basis.

  • 4
    $\begingroup$ You are conflating a polynomial $p$ with the equation $p(x) = 0$. In particular, a polynomial $p$ is zero if and only if all its coefficients are zero. $\endgroup$ Aug 9, 2021 at 16:22
  • $\begingroup$ @JoseAvilez you are saying that $3ax^2+2bx+c=0 \implies a=b=c=0$ right? so we get only $d$. Why then my solution ain't right, and I get a different solution? $\endgroup$
    – Chopin
    Aug 9, 2021 at 16:38
  • $\begingroup$ No. I'm saying that if $3ax^2 + 2bx + c $ is the zero polynomial, then $a=b=c=0$. $\endgroup$ Aug 9, 2021 at 16:40
  • $\begingroup$ @JoseAvilez Oh, I see. So since all of them are zero, what I am doing is just playing with zeros, and therefore I allegedly get different solution. Right? $\endgroup$
    – Chopin
    Aug 9, 2021 at 16:42

1 Answer 1


You can see the problem in the line $$ \left\{ax^{3} +bx^{2} +cx+d\Bigl| c=-3ax^{2} -2bx, \,a,b,c,d\in \mathbb{R}\right\}. $$

On one hand, you write $c \in \mathbb{R}$ (which is true as $a,b,c,d$ are the coefficients of the polynomial $p(x)$) but on the other hand, you write $$ (1) \qquad\qquad c =-3ax^{2} -2bx $$ which means that $c \in P_2[\mathbb{R}]$. How can it be?

Strictly speaking, writing $c = -3ax^2 - 2bx$ is a wrong and a bad idea so I do not recommend it. However, if you insist on doing it then there is a way to make it rigorous. The only way for equality $(1)$ to make sense is to think of of $c$ both as a real number and as a polynomial in $P_2[\mathbb{R}]$. This means that you need to identify $c \in \mathbb{R}$ with a constant polynomial $$ \hat{c} = c \cdot 1 + 0 \cdot x + 0 \cdot x^2$$ and then the equation $ c = -3ax^2 - 2bx$ actually means $$ \hat{c} = -3ax^2 - 2bx \iff 0 \cdot x^2 + 0 \cdot x + c \cdot 1 = (-3a) \cdot x^2 + (-2)\cdot bx + 0 \cdot 1. $$ Using the definition of equality of polynomials and comparing coefficients, this means that $-3a = 0, -2b = 0$ and $c = 0$ so $a = b = c = 0$ and you arrive to the correct conclusion.

  • $\begingroup$ Got it all!!! Thank you for this clear and wonderful answer. $\endgroup$
    – Chopin
    Aug 9, 2021 at 16:57
  • $\begingroup$ @Chopin: Sure. To hammer the point, when you differentiate $ax^3 + bx^2 + cx + d$, you treat $c$ as a real number and not a polynomial, right? For otherwise, the derivative would be $3ax^2 + 2bx + c'(x) \cdot x + c$. Then, when you suddenly replace the role of $c$ and go back the original expression, you actually used the "wrong formula" to differentiate! $\endgroup$
    – levap
    Aug 9, 2021 at 17:04
  • $\begingroup$ Right. And by the way, if I wanted to prove the Lamma in which if a polynomial is zero for all $x$ then the coefficients are zeros, then I could just doing the following: for example for $p_2[x]$ we take $ax^2+bx+c=0$ and then $c=-ax^2-bx$ but it is a contradiction unless $a=b=0$. So is it right? for one direction of the proof of course. $\endgroup$
    – Chopin
    Aug 9, 2021 at 17:11
  • $\begingroup$ @Chopin: Actually, there are two distinct notions which are commonly called "polynomials". One is a "polynomial function" which means a function of the form (say) $ax^2 + bx + c$ and the other is an "algebraic polynomial" which is just an "expression" of the form $ax^2 + bx + c$. I have worked with the algebraic polynomial definition and in that case, the fact that if $ax^2 + bx + c = 0$ then $a = b=c=0$ is an immediate consequence of the definition. However, with the "polynomial function" definition, the meaning of $ax^2 + bx + c = 0$ is that you have an equality for all $x \in \mathbb{R}$. $\endgroup$
    – levap
    Aug 9, 2021 at 17:15
  • 1
    $\begingroup$ To deduce that $a = b = c = 0$, you actually need to work a little. Plug in $x = 0$ to get $c = 0$. Then $ax^2 + bx = 0$. Now plug in $x = 1$ to get $a + b = 0$ and $x = -1$ to get $a - b = 0$ which implies that $a = b = 0$. Writing $c = -ax^2 - bx$ and saying that this a contradiction is not rigorous enough because you need to justify why it is not possible for $c = -ax^2 - bx$ to be true for all $x \in \mathbb{R}$. This is intuitively clear but the formal justification is done by plugging in various $x$'s and deducing the result. $\endgroup$
    – levap
    Aug 9, 2021 at 17:17

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