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.
Best Answer
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.