# Prove or Disprove the existence of a basis

I'm asked to prove or disprove the existence of a basis $(p_0,p_1,p_2,p_3)$ of $F(t)(3)$ (Polynomials of degree at most 3) such that each of the polynomials $p_0,p_1,p_2,p_3$ satisfies the equation $$tp''(t)+3p'(t)=0.$$

We're not supposed to know how to solve a differential equation to solve this problem. I've to confess that I'm a bit lost, can anyone give me a hint ?

Thanks

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"We're not supposed to know how to solve a differential equation to solve this problem." But you know the theory of linear ODEs? – Martín-Blas Pérez Pinilla Jan 31 '14 at 22:28
Looking at the different answers I just received, it seems that not. I'm going to work on my side in order to take advantage of all these hints. – Alexia Azoulay Jan 31 '14 at 22:36

Your condition is actually just requiring $p_0, p_1, p_2,$ and $p_3$ to be in the kernel of a linear map over the vector space of polynomials of degree at most $3$--the linear map is $L(p) = t\,p''(t) + 3\,p'(t)$ (You should prove that it's linear!). You can write this out in matrix form (if you wish), using the standard bases $\{ 1, t, t^2, t^3 \}.$ The question then becomes: Does the kernel of this map have dimension $4$?

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Thanks a lot for your help, it's very clear. I found the kernel of this map to have dimension different of 4 ; therefore it couldn't represent a basis of F3(t) ? I'm very unsure though. – Alexia Azoulay Jan 31 '14 at 22:45
No, because a basis would have to have 4 vectors (polynomials in this case) because the dimension of your vector space is dimension 4. – breeden Jan 31 '14 at 22:51
@Alexia you're right. $F3(t)$ is a four dimensional space, but the solution space to the DE is a two dimensional space (the nullspace or kernel of a certain linear operator). Since it's two dimensional, any linearly independent set of solutions can have at most two elements. No set of four solutions could be linearly independent, and so could not form a basis for anything. – Unwisdom Jan 31 '14 at 22:58
@Unwisdom - thanks for adding some clarifying remarks. When I said "No," I did mean "No, therefore it couldn't represent a basis of F3(t)". I can see how my "No" could be a little unclear. – breeden Jan 31 '14 at 22:59

Well, what you have is a second order linear differential equation. So, without solving it, what is the dimension of the solution space of that DE?

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I'm pleased you asked this, since it gives me a chance to link to my own answer to the question "what was the first bit of mathematics that made you realize that math is beautiful", which would otherwise likely never be read by anyone. It is directly relevant to your question. math.stackexchange.com/a/657639/124220 – Unwisdom Jan 31 '14 at 22:47

Idea: $${p''(t)\over p'(t)}={-3\over t}.$$ Can you continue now?

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Any linear combination of solutions to this differential equation is again a solution. So if all vectors of a basis are solutions then all vectors of the whole space are. Are they?

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