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The set $\mathbb R$[x] of polynomials in $x$ with real coefficients is a real vector space. The map $\delta$ : $\mathbb R$[x] $\to$ $\mathbb R$[x] is defined as follows: for $f$ $\in$ $\mathbb R$[x], define $\delta$($f$) = $df/dx$. That is, $\delta$($f$) is the derivative of $f$ with respect to $x$.

a) Show that $\delta$ is a linear transformation.

b) Find ker $\delta$ and im $\delta$

c) Is $\delta$ injective? Is $\delta$ surjective? Is $\delta$ an isomorphism?

For part a)

Checking for addition:

Take two functions $f$ and $g$ from $\mathbb R$[x].

$\delta$$(f + g)$ = $d/dx(f+g)$ = $df/dx$ + $dg/dx$ = $\delta$$(f)$ + $\delta$$(g)$.

Checking scalar multiplication:

Take a function $f$ from $\mathbb R$[x] and let $\lambda$ be a scalar.

$\delta$$(\lambda f)$ = $d(\lambda f)/dx$ = $\lambda$$df/dx$ = $\lambda$$\delta$$(f)$

Both addition and scalar multiplication are satisfied hence $\delta$$(f)$ is a linear transformation.

I am stuck on part b). I think that ker $\delta$$(f)$ = {$f$$\in$$\mathbb R$[x] : $\delta$$(f)$ = $0$}.

$\delta$$(f)$ = $dy/dx$ = $0$ if and only if $f$ is constant. Therefore, the kernal of $\delta$$(f)$ is all polynomials of degree $1$. This is one-dimensional space spanned by $\lambda$ where $\lambda$$\in$$\mathbb R$.

Is this correct? Since ker ($f$) is not {$0$} then $\delta$$(f)$ cannot be injective and thus $\delta$$(f)$ is not an isomorphism (part c)).

I am wondering if what I have done so far is correct and how to complete the remainder of the question i.e. what is im $\delta$$(f)$ and is it surjective?

serveoverice
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    For b), you have correctly stated the definition of the kernel. However, which functions have that property? – Ben Grossmann Mar 20 '16 at 14:22
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    Isn't it all polynomials of degree $1$? – serveoverice Mar 20 '16 at 14:23
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    Your argument about the kernel is correct. The kernel is given by constant polynomial and therefore is a 1 dimensional space generated by any not null constant polynomial. Since the kernel is not ${0}$ you have $\delta$ is not isomorphism. The map is surjective because if $g$ is a polynomial the you can consider $f(x) = \int_{0}^{x} g(t)dt$. Just note that $\delta (f) = g$. – Hugo Mar 20 '16 at 14:25
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    Thanks Hugo. Any idea on the image? I'm not sure I understand this concept properly. – serveoverice Mar 20 '16 at 14:27
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    The image is $\mathbb R[x]$. The polynomial $g$ is on the image of $\delta$ when there is a $f\in \mathbb R [x]$ such that $\delta(f)=g$. Note that this statement is true for every $g\in \mathbb R [x]$ because you can take $f(x)=\int_{0}^{x} g(t)dt$. Then $\delta (f) = \frac{d}{dx}\int_{0}^{x} g(t)dt=g(x)$. – Hugo Mar 20 '16 at 14:31
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    Thanks Hugo, much appreciated. – serveoverice Mar 20 '16 at 14:36
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    You've done everything correctly as far as I can tell, just note that the degree of a constant polynomial is $0$ not $1$. – K.Power Mar 20 '16 at 14:44
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    Okay, thanks K.Power. – serveoverice Mar 20 '16 at 14:47

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