1
$\begingroup$

Consider the map $L\colon P_n(\mathbb{R}) \rightarrow P_n(\mathbb{R})$, $L(P(x)) = P(x) - xP'(x),$ where $P_n(\mathbb{R})$ is the vector space of all polynomials of degree at most n with coefficients in $\mathbb{R}$.

a) Determine the kernel of $L$. What is its dimension?

b) Determine a basis for the image of $L$.

I'm not confident in getting the kernel. I know I need $L(v) = 0$, but I always feel like I'm missing something. Any help is appreciated.

$\endgroup$
1
  • $\begingroup$ "I always feel like I'm missing something" is sort of vague. Can you give more detail? $\endgroup$ Commented Feb 17, 2013 at 23:18

4 Answers 4

1
$\begingroup$

A visual way to do that is to compute the image of the canonical basis $\{1,x,\ldots,x^n\}$.

This is $\{1,0,-x^2,-2x^3,\ldots,-(n-1)x^n\}$.

If you want, now you can write down the matrix of $L$ with respect to this basis.

Either way, you should easily see that $$ \mbox{Ker }L=\mathbb{R}x \quad \mbox{and}\quad \mbox{Im }L=\mbox{span}\{1,x^2,\ldots,x^n\}. $$

$\endgroup$
2
  • $\begingroup$ ok, thats what I thought for the kernel and image. However will the span{1,x^2,...} work for the basis of the image? Bases and spanning sets usually through me off. I'm not too great with terminology $\endgroup$
    – Moses G
    Commented Feb 17, 2013 at 23:37
  • $\begingroup$ @MosesG Yes, this set is linearly independent (it is a subfamily of the canonical basis). So it is a basis of its span. $\endgroup$
    – Julien
    Commented Feb 17, 2013 at 23:40
1
$\begingroup$

You do indeed need all polynomials $p(x)$ such that $L(p(x))=0$, that is, $p(x)-xp^\prime(x)=0$.

If we denote $p(x)=a_0+a_1x+\ldots+a_nx^n$, then $xp^\prime(x)=a_1x+2a_2x^2+\ldots+ka_kx^k+\ldots+na_nx^n$.

To get $p(x)-xp^\prime(x)=0$, you need that for all $k\in\{0,\ldots,n\}$, $a_k-ka_k=0$, or in other words, $(k-1)a_k=0$, so for all $k\ne1$, $a_k=0$, and $a_1$ can be whatever you want.

So $\ker L = \{ax | a\in \mathbb{R}\}$.

$\endgroup$
2
  • $\begingroup$ You want $p-xp'=0$ in the third paragraph. $\endgroup$
    – Pedro
    Commented Feb 17, 2013 at 22:18
  • $\begingroup$ Corrected, thank you. $\endgroup$ Commented Feb 17, 2013 at 22:22
0
$\begingroup$

Every element of $P_n(\mathbb R)$ can be written in the form $$f = a_0 + a_1x + \cdots + a_nx^n$$ for some choice of $a_i \in \mathbb R$. Taking the derivative and multiplying by $x$ gives $$a_1x + 2a_2x^2 + 3a_3x^3 + \cdots + na_nx^n$$ therefore $$L(f) = a_0 + 0\cdot a_1x - a_2x^2 - 2a_3x^3 - \cdots - (n-1)a_nx^n$$

This is zero if all the coefficients are zero, so set them all equal to zero. You now have a bunch of equations involving the $a_i$. Try and solve them and plug the answers back into your expression for $f$. This will tell you what elements of the kernel look like and you can work from there.

$\endgroup$
0
$\begingroup$

Hint: Let $P(x) = a_nx^n + \dots + a_0$. Then $xP'(x) = na_nx^n + (n-1)a_{n-1}x^{n-1} + \dots a_1x$. Then $P(x) - xP'(x) = 0 \Rightarrow a_i = ia_i$ for all $i = 1, \dots n$, and $a_0 = 0$. Thus, $a_n = a_{n-1} = \dots = a_2 = 0$, and $a_0 = 0$. This means that $P(x) = a_1x$.

Another way is the following: You have a basis for $P_n(x)$ which is $\{1,x,x^2,\dots, x^n\}$. You see that $L(x) = 0$, so the kernel has dimension at least $1$. Similarly, \begin{equation} L(1) = 1, L(x^2) = -x^2, L(x^3) = -2x^3, \dots, L(x^n) = -(n-1)x^n. \end{equation} We know that if a set of vectors $\{v_,\dots,v_k\}$ is linearly independent, then the set $\{a_1v_1, \dots a_kv_k\}$ is also linearly independent for $a_j \in \mathbb{R} - 0$. Thus, $\{1, -x^2, -2x^3, \dots, -(n-1)x^n\}$ is linearly independent. So, $\dim(im(L)) \geq n-1$. Since $\dim(ker(L)) \geq 1$, by rank nullity, we must have $\dim(im(L)) = n-1$ and $\dim(ker(L)) = 1$. We already know a non-zero vector in the kernel, namely $x$. Then it should be easy to say what the rest of the kernel is. We also obtained a basis for the image by the way!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .