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In my textbook is written that every linear mapping $ f: V \to V'$ satisfies $f(0)=0$. But what about the mappings between the polynomials like $p(x) = ax^2 + bx + c$, where $c$ is nonzero? Elementary probably, but fundamental.

3 Answers 3

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There is a null polynomial $p(x) = 0$. What you mean is an affine space, not a linear space.

About the property:

Every vector space has a null vector $\vec{0}$ so for a linear map $f$ we have $$ f(\vec{0}) = f(0 \, \vec{0}) = 0 f(\vec{0}) = \vec{0} $$

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    So all I need is just to have guaranteed that in both of my mapped vector spaces exists a null polynomial? But in the case of $p(x) = ax^2+bx+c$, ... $p(0) = a0 + b0 + c = 0$, so $c$ must be zero here. The polynomials create a vector space, but in general mappings I can have the transformation where $c$ is nonzero, but not if $x$ is zero, because I have to satisfy that $p(0) = 0$. – 2017-02-13
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    @Leif A zero always exists in a vector space, by definition. So you don't have to verify that. You do have to verify that the set you are mapping from is a vector space, though. – 2017-02-13
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    @Leif It looks like you might be a bit confused here. The zero polynomial is the one that evaluates to $0$ for *all* values of $x$, i.e., the constant $0$. Simply having $p(0)=0$ doesn't make $p$ the zero of the vector space. – 2017-02-13
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    A set of polynomials can be a vector space only if it contains the null polynomial $p(x)=0$ and, in your case, this is the polynomial with $a=b=c=0$. Remember that the null polynomial is such that $p(x)=0$ for all values of $x$, not simply for $x=0$. – 2017-02-13
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The set of polynomials $ax^2+bx+c$ with $c\ne 0$ t is not a vector space (at least with the usual operations), so we cannot speak of linear mapping on this set.

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    I thought that vector spaces with polynomials are ok with the constants. I can have the mapping, where $c$ is nonzero, but for $p(0) = a0 + b0 + c = 0$, $c$ must be a zero, but if I am not mapping the null polynomial, that constant can be nonzero, or not? – 2017-02-13
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    The null polynomial has $a=b=c=0$ and is null for all values of $x$. See my other comment. – 2017-02-13
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You are getting confused between $f$ and $p$.

$f$ is a function that applies to the coefficients of the polynomial and returns new coefficients.

For clarity, we can write

$$f(a,b,c)=(a',b',c')$$

and a necessary condition for linearity is

$$f(0,0,0)=(0,0,0).$$

For instance,

$$f(a,b,c)=(2a,b+c,c-a),$$ also understood as

$$f(ax^2+bx+c)=2ax^2+(b+c)x+(c-a)$$

may be linear because

$$f(0x^2+0x+0)=0x^2+0x+0.$$

This has nothing to do with the exaluation of the polynomial at $x=0$, and mandatorily involves a polynomial such that $c=0$.

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    Alternatively, $f(0) = f(0+0) = f(0) + f(0)$, which implies that $f(0) = 0$. – 2017-02-13
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    I have completely rewritten my answer. – 2017-02-13
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    Perfect, thank you very much, I was really confusing it. – 2017-02-13