Let [itex]V= \mathbb{R}_3[x][/itex] be the vector space of polynomials with real coefficients with degree at most 3 and let [itex]D:V\to V[/itex] be the linear operator of taking derivatives, [itex]D(f)=f'[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

I'm trying to check the Rank-nullity theorem for this example but it doesn't seem to hold:

Since [itex]D[/itex] is not injective (e.g. images of [itex]x^2[/itex] and [itex]x^2+1[/itex] are both [itex]2x[/itex]), [itex]\text{Ker}(D)\neq \{ \bf{0} \}[/itex]. (The kernel is actually all the constant polynomials).

Since [itex]D[/itex] is surjective (as every polynomial is the derivative of some other polynomial), [itex]\text{Im}(D)=V=\mathbb{R}_3[x][/itex].

Now the Rank-nullity formula is:

[itex]\text{dim}(\text{Ker}(D)) + \text{dim}(\text{Im}(D)) = \text{dim}(V)[/itex].

But [itex]\text{dim}(\text{Im}(D)) = \text{dim}(\mathbb{R}_3[x]) = 4 = \text{dim}(V)[/itex]

so if the Rank-nullity formula holds this would imply [itex]\text{dim}(\text{Ker}(D))=0[/itex] but it doesn't as [itex]D[/itex] is not injective, so what's going on?

EDIT: D is not surjective! How do I find a basis for the Kernel and Image?

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# Let [itex]V= \mathbb{R}_3[x][/itex] be the vector space of polynomials

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