Dimension of vector space of polynomials I read on vector spaces lately and then I did a thought experiment. As for a vector in R3 the dimension is 3 (maybe you could give me a proof?), i asked myself what would the dimension be of a vector space of polynomials with degree <= k? I currently think it is infinite as you can always add another polynomial that is not expressible as a linear sum of the set, but I do not have any solid proof for this statement.
 A: Any quadratic polynomial $ax^2+bx+c$ is obviously a linear combination of the three polynomials $x^2$, $x$ and $1$, so that the space of polynomials of degree $\le2$ is at most of dimension $3$.
You can generalize to any degree.
A: The set $\{1, x, x^{2}...,x^{k}\}$ form a basis of the vector space of all polynomials of degree $\leq k$ over some field. Every polynomial will be in some linear combination of these vectors. Also it is not difficult to show that the above set is linear independent. So dimension of the vector space is $k+1$. Your vector space has infinite polynomials but every polynomial has degree $\leq k$ and so is in the linear span of the set $\{1, x, x^{2}...,x^{k}\}$.
$$OR$$
Basis is maximal linear independent set or minimal generating set. Since every polynomial is of degree $\leq k,$ set $\{1, x, x^{2}...,x^{k}\}$ is a minimal generating set or a basis. Now one can say by definition of basis that any $k+2$ vectors form a linear dependent set.
A: Recall that the definition of a basis of a vector space is a set of vectors such that they are linearly independent, i.e if any linear combination of the basis vectors is zero then the coefficients of each basis vector must be zero and that they span the vector space i.e any vector can be written as a linear combination of the basis. So a basis for $R^3$ is simply the set {(1,0,0),(0,1,0),(0,0,1)}. Of course there are infinitely many possible bases for a vector space, {(2,0,0),(0,1,0),(0,0,1)} is another basis for $R^3$. Similarly {$1,x,x^2,x^3,...,x^{k}$} is a basis for the set of polynomials of degree k. You can easily see that this set satisfies the above 2 properties to be a basis.
I would have commented above but I don't have enough reputation. To answer your question about why the dimension of the vector space has to be unique, see https://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces
