Dimension of a system of equations and the number of equations required to get a unique solution I'm a bit confused about how I can get how many equations I need to get a unique solution for a system of equations. For example, if I have a polynomial function of x at degree n, how many equations do I need to get a unique solution for f(x)?
To my understanding, f(x)=x+x^2+x^3....+x^n. The function is n-dimensional, so we need n equations. Is that correct?
What if we have a function f(x) = x^2*y + y^2*x, how many equations do I need for that? Thank you!
 A: You have an important misunderstanding here: A polynomial is not generally of the form $x+x^2+x^3+\cdots+x^n$ but it may be allowed to have a constant term and coefficients; in other words for any polynomial $f$ there are $n+1$ numbers $a_0,a_1,a_2,\dots,a_n$ such that
$$f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots + a_nx^n$$
Now, if you know that a polynomial goes through some points, say, for the sake of argument, $(x,f(x))=(1,0)$ and $(3,5)$, then you can get some equations:
$$0 = a_0 + a_1(1) + a_2(1)^2 + a_3(1)^3 + \cdots + a_n(1)^n$$
$$5 = a_0 + a_1(3) + a_2(3)^2 + a_3(3)^3 + \cdots + a_n(3)^n$$
Now, in these equations, what is playing the role of the variables that you need to solve for?
A: The answer is $n+1$. A good way to remember it is to assume you're trying to make the polynomial take on the value zero at several points. Suppose the polynomial is of degree $n$. Then it can have at most $n$ zeros before it must be identically the zero function. So $n+1$ values forces it to be a specific function. This is a specific case of the more general rule that there is a unique polynomial of degree $\leq n$ that takes on the values $p(a_1)=b_1, \dotsc, p(a_{n+1})=b_{n+1}$, where $a_i \neq a_j$ for $i\neq j$.
A fun linear algebra way to look at this is that the polynomials of degree $\leq n$ are a vector space of dimension $n+1$, and given points $a_1, \dotsc, a_{n+1}$, we can construct a basis $v_1(x), \dotsc, v_{n+1}(x)$(the Lagrange polynomials) by
$$v_i(x) = \frac {(x-a_1)\dotsb(x-a_{i-1})(x-a_{i+1})\dotsb(x-a_{n+1})} {(a_i - a_1)\dotsb(a_i - a_{i-1})(a_i-a_{i+1})\dotsb(a_i - a_{n+1})}.$$
Notice $v_i(a_j) = \begin{cases}0 &\text{if}j\neq i\\1 &\text{if} j=i.\end{cases}$
