4
$\begingroup$

I have the following linear map:

$$T:\mathcal{P}^2 \to \Bbb{R}^2 $$

where $\mathcal{P}^2$ denotes the vector space of polynomials with real coefficients having degree at most $2$

$ T $ is defined by $$T(ax^2+bx+c)=[a+b , b−c]^t$$


I do not know how to prove that $T$ is surjective.

I know its not injective.

Yet,i do not now how to formally show that is a surjection.

I tried following the answer at How to show that a linear map is surjective?

Using the formula described:

  • $\dim(V) = 2$,
  • $\dim(\operatorname{range}T)=1$

According to the formula "will be surjective if $\dim V= \dim \operatorname{range} T$" this map would not be surjective.

Yet, in the website i took this exercise from, it says this map is a surjective.

So, i am wondering what i did wrong, and if there are better ways to show that this map is a surjective.

I came across on this exercise on this website.

$\endgroup$
0

4 Answers 4

3
$\begingroup$

$T:\mathcal{P}^2\to \Bbb{R}^2$ defined by $$T(ax^2+bx+c) =\begin{pmatrix}a+b\\b-c\end{pmatrix}$$

$\ker{N}(T)=\{p\in\mathcal{P}^2 :T(p) =0\}$

To find the $\ker T$ we have to solve a system of two equations with three unknowns.

$$a+b=0\tag 1$$

$$b-c=0\tag 2$$

We get $c=b=-a$ and $p(x)=ax^2+bx+c\in\ker T$

$\implies b=-a, c=b=-a$

Hence null space basis is $\{\begin{pmatrix}1\\-1\\-1\end{pmatrix}\}$

Hence $\dim\ker T=1$ and by Rank-Nullity theorem :

$$\dim \ker T +\dim \operatorname{range} T =\dim \mathcal{P}^2$$

Hence $\operatorname{range} T=3-1=2=\dim(\Bbb{R}^2)$.

Hence $\operatorname{range} T=\Bbb{R}^2$ implies $T$ is surjective.

Note: $\dim(\mathcal{P}^2) =3 $ as basis $\{1, x, x^2\}$.

$\endgroup$
3
$\begingroup$

For any $(p,q)\in\mathbb R^2$, $T(px^2-q)=(p+0,0--q)=(p,q)$. Since this is true for any pair, $T$ is surjective.

$T(x^2+5x-3)=T(2x^2+4x+2)=(6,2)$ so $T$ is not injective.

$\endgroup$
2
$\begingroup$

I recently make this figure to explain the concepts of injectivity and surjectivity and some related properties related to the rank and left/right inverses:

Linear mappings

Although it is stated for matrices, it is valid for linear maps between spaces of finite dimension.

Your case corresponds to the green area in the above figure. Using the standard basis of $\mathsf{P}_2({\rm I\!R})$, that is, $\{1, x, x^2\}$, we have that your linear map is described by the matrix $M_T\in{\rm I\!R}^{2\times 3}$ $$ M_T = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & -1 \end{bmatrix}, $$ and it easy to see that $\mathrm{rank} M_T = 2$, which proves that $M_T$ (and $T$) is surjective.

$\endgroup$
0
$\begingroup$

@Ayla To fit more precisely to your 2 questions:

  • "So, i am wondering what i did wrong": it is your computation of $\dim(\operatorname{range}(T))$ which is wrong. You claim (without proof) it is equal to $1$, whereas the correct result is $2$. You don't tell us how your got this wrong $1$ but I suppose it was by an incorrect computation of $\dim(\ker(T))$ or incorrect application of the rank-nullity theorem (for a correct version, see Sourav Ghosh's answer).
  • "and if there are better ways to show that this map is a surjective": a direct computation is good enough. $$T(ax^2+bx+c) =\begin{pmatrix}u\\v\end{pmatrix}\Leftrightarrow\begin{cases}a+b&=u\\b-v&=c\end{cases}\Leftrightarrow\begin{cases}a&=u-b\\c&=-v+b\end{cases}$$ so any $\begin{pmatrix}u\\v\end{pmatrix}\in\mathbb R^2$ is the image (by $T$) of many polynomials, e.g. (choosing $b=0$) $$\begin{pmatrix}u\\v\end{pmatrix}=T(ux^2-v).$$
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.