Finding bases for vector spaces with two constraints

Question

I'm having problems understanding this question. Compute the dimension and find bases for the vector space of polynomials $p(x)$ of degree 4 which have the property that $p(2) = 0$ and $p(3) = 0$.

Right now, I have tried substituting x = 2 and x = 3 into the polynomial equation $a + bx + cx^2 + dx^3 + ex^4$, then I obtained two equations. I put these two equations into a matrix and solved for $a$ and $b$ in terms of the other three variables, $c, d, e$. I then substituted these variables into the equation, thus obtaining an equation of $p(x)$ in terms of $c, d, e$.

I ended up with the bases $(6-5x+x^2), (30-19x+x^3), (114-65x+x^4)$. However I'm not sure if this is correct - seems like they are not independent (could not row reduce to 1s on diagonal..)

Does anyone have an idea of how to solve this question correctly? Would really appreciate it!

Answer 1

You have to use some arithmetic of polynomials: $p(2)=0$ means $p(x)$ is divisible by $x-2$. Similarly $p(x)$ is divisible by $x-3$. As $x-2$ and $x-3$ are coprime, $p(x) $ is divisiible by $(x-2)(x-3)$ by Gauß's lemma.

So we can write: $$p(x)=(x-2)(x-3)q(x),\quad \deg q=2,$$ so that the vector space of polynomials $p$ such that $p(2)=p(3)=0$ is isomorphic to the vector space of polynomials of degree at most $2$ through the mapping $p\longmapsto q(x)=\dfrac{p(x)}{(x-2)(x-3)}$.

This shows the vector space has dimension $3$. Since the reciprocal isomorphism is $q(x)\longmapsto (x-2)(x-3)q(x)$ and we know a basis for the polynomials of degree at most $2$, a basis for the vector space is, e.g.: $$\{(x-2)(x-3), x(x-2)(x-3),x^2(x-2)(x-3)\}.$$

Answer 2

They are indeed linearly independent, since each has a term of a degree not shared by any other: only the first one has an $x^2$ term, only the second has an $x^3$ term, and only the third has an $x^4$ term.

Observe that if you arrange them into a matrix as $$\begin{bmatrix} 1 & 0 & 0 & -65 & 114\\ 0& 1 & 0 & -19 & 30 \\ 0 & 0 & 1 & -5 & 6 \\ \end{bmatrix}$$

you will see that you already have row-reduced them. The fact that the right-hand columns are not zeros is just a consequence of the fact that you are in a 5-dimensional vector space but have only 3 vectors.