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$ax^2 + bx + c = 0$. How many linearly independent vectors $(a,b,c)$ are there that will yield the same root $r$? Example, if $r = 0$, some independent vectors are $(1,0,0)$ and $(1,1,0)$. How do things change if we only consider real numbers vs complex numbers. What if the root is complex but the coefficients are real. Where can I learn more?

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  • $\begingroup$ Quick guess, not an answer since I haven't thought it through. I think you can assume the root is $0$, with no loss of generality. Then you've solved the problem - the rank is $2$. If I'm right, answer your own question. Warning: I may be all wrong. $\endgroup$ – Ethan Bolker May 3 '17 at 23:43
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For fixed $ x $, the map $ f : K^3 \to K $ for a field $ K $ defined by $ f: (a, b, c) \to ax^2 + bx + c $ is a surjective $ K $-linear map, therefore its kernel is a $ 2 $-dimensional subspace of $ K^3 $ by the rank-nullity theorem. In other words, for any $ x $, the subspace of vectors $ (a, b, c) \in K^3 $ such that $ ax^2 + bx + c = 0 $ is $ 2 $-dimensional, i.e it is spanned by two linearly independent vectors. Note that this result is independent of the choice of the field $ K $, therefore the result remains the same over $ \mathbb R $ or over $ \mathbb C $.

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If $a \neq 0$, then $$a x^2 + b x + c = a (x-r)(x-p) = ax^2 - (a r + ap ) x + a r p$$

where $p$ is the other root. Since $a$ and $p$ can be freely changed without affecting that $x=r$ is a root, we have two degrees of freedom.

If we do allow the possibility for $a=0$, then

$$0 x^2 + b x + c = b (x + c/b) = b (x-r)$$

which leaves one degree of freedom. As such, there are three linearly independent vectors that yield the same root $r$ or two if we require $a \neq 0$. Note that the space of solutions isn't a vector subspace---it does not include the zero vector.

This analysis is indifferent to all coefficients being real or complex (do keep in mind that one complex dimension is isomorphic to two real dimensions).

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