Dimension of the vector space of homogeneous polynomials

Added later: Except the formula is correct, as shown in the comments below; I had mis-read it; when $d = 2$ and you have 3 variables, $n = 2$, since the variable indexing starts at zero. And 4-choose-2 is in fact six, as expected.

[what follows is incorrect]

A nice basis for that space consists of all monomials in the $n$ variables with total degree $d$.

Wait...what about degree $d = 2$ and $n = 3$ variables. Listing the basis elements, I see $$ x^2, y^2, z^2, xy, xz, yz, $$ which is only $6$ dimensions, but 5 choose 3 is 10. Seems as if there might be a mistake in your formula, or perhaps I'm not understanding how it's supposed to be applied. (See below for resolution of this apparent contradiction.)

I think (or thought) perhaps your formula is incorrect.

Tags:

Linear Algebra

Polynomials

Vector Spaces

Related

I found this odd relationship, $x^2 = \sum_\limits{k = 0}^{x-1} (2k + 1)$. A family has three children. What is the probability that at least one of them is a boy? Show that if G is a simple graph with at least 4 vertices and 2n-3 edges, it must have two cycles of the same length. Prove that any palindrome with an even number of digits is divisible by 11 Integral $ \int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x $ Jacobian criterion for projective varieties How to evaluate $\int_0^\infty\operatorname{erfc}^n x\ \mathrm dx$? Another Epsilon-N Limit Proof Question I Need Help Understanding Quantifier Elimination Is a morphism of reduced schemes over an algebraically closed field determined by its values on closed points? How can we describe the graph of $x^x$ for negative values? How many squares actually ARE in this picture? Is this a trick question with no right answer?

Recent Posts