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I have here a linear transformation $T : P_3(\mathbb{R})\rightarrow P_3(\mathbb{R}) $ defined by:

$ T(at^3 + bt^2 + ct + d) = (a-b)t^3 + (c-d)t $

I'm very very new in this subject and I'm not going well with polynomials. I need find the $ Kernel $ and the $ Image $ of the transformation. Look what I've been thinking:

$Ker(T) = \{ T(p) = 0 / p \in P_3\} $

$ T(at^3 + bt^2 + ct + d) = (a-b)t^3 + (c-d)t = 0 $

$(a-b) = 0 \ ;\ \ (c-d) = 0 \ ;\ \ a = b \ ; \ \ c = d $

$ Ker(T) = \{ at^3 + at^2 + ct +c\ /\ a,c \in \mathbb{R} \} $

And what about the $ Image $? I know that $Im(T) = \{ T(p) / p \in P_3 \}$, but how can I show it? And how can I test if a polynomial such as $ p(t) = t^3 + t^2 + t -1 \in Im(t)$?

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  • $\begingroup$ Can you write $t^3 + t^2 + t - 1$ as $(a - b) t^3 + (c - d) t$? $\endgroup$ – user66081 Feb 10 '15 at 12:00
  • $\begingroup$ When are two polynomials equal? $\endgroup$ – Tomás Feb 10 '15 at 12:06
  • $\begingroup$ I can't. So isn't It in Im(T)? $\endgroup$ – João Paulo Feb 10 '15 at 12:07
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    $\begingroup$ Right. So that should give you an idea what is in Im(T) $\endgroup$ – user66081 Feb 10 '15 at 12:08
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The kernel is correct. Additionally, since the kernel depends on only two coefficients $a$ and $c$, it has dimension 2.


For the image:

Take any polynomial $p(t)=At^3+Bt^2+Ct+E$.

The question now is: How do $A,B,C,E$ have to look for there to exist some $a,b,c,d$ such that $T(at^3+bt^2+ct+d)=p(t)$?

The question is equivalent to solving for $A, B, C, E$ in the equation: $(a-b)t^3+0t^2+(c-d)t+0=At^3+Bt^2+Ct+E$.

We now have:
$A=a-b$,
$B=0$,
$C=c-d$
$E=0$

We can take:
$a=A$,
$b=0$,
$c=C$
$d=0$

Consequently, $p$ is in the image, iff $B=0=E$. The image, then, is: \begin{align*} \mbox{Im}(T)=\{At^3+Ct\ |\ A,C\in\mathbb R\}. \end{align*}

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  • $\begingroup$ The Image also has dim = 2, so $ dim(Ker) + dim(Im) = dim(P_3) = 2 + 2 = 4 $... I think I got it... thanks!! $\endgroup$ – João Paulo Feb 10 '15 at 12:12
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We can set up the matrix of the linear transformation $T:P_3(\mathbb{R})\rightarrow P_3(\mathbb{R})$, then find its null space and column space, respectively. First, if we agree to represent the third-order polynomial $P_3=at^3 + bt^2 + ct + d$ by the column vector $\begin{pmatrix}a &b& c& d\end{pmatrix}^T$, then $$T=\begin{pmatrix} 1 & -1 & 0 &0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & -1\\ 0 & 0 & 0 & 0 \end{pmatrix}.$$

It is obvious that the pivots of $T$ is on the first and the third columns, so the kernel of T (i.e., the null space) is $$\mathrm{ker}\, T=\mathrm{span}\{\begin{pmatrix} 1\\1\\0\\0 \end{pmatrix},\begin{pmatrix} 0\\0\\1\\1 \end{pmatrix}\},$$ and the image of $T$ is spanned by the pivot columns of $T$: $$\mathrm{im}\, T=\mathrm{span}\{\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix},\begin{pmatrix} 0\\0\\1\\0 \end{pmatrix}\}.$$

The answers are identical to what have been given. This is just to show we can get the answers by a slightly different approach.

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