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Let $V$ be a vector space of homogeneous polynomials in 3 variables $x_1, x_2$ and $x_3$ over $\mathbb{R}$.

What is $\dim V$?

I think it will be some expression in terms of $d$ but I am not sure how to find the answer

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thinker
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    you perhaps mean degree at most $d$, otherwise this cannot be vector space, as you cannot have a zero vector. – ugur efem May 01 '16 at 11:07
  • No, I am only considering degree equal to $d$ – thinker May 01 '16 at 11:15
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    Well in that case $V$ is not a vector space. Look at the polynomials $x_1x_2x_3$ and $-x_1x_2x_3 + x_1^2$. Both are of degree $3$ but their sum is not – ugur efem May 01 '16 at 11:19
  • If you see my original post, I have attached the corresponding question-it says vector space of degree d – thinker May 01 '16 at 11:27
  • ah it says homogeneous! That changes everything now – ugur efem May 01 '16 at 11:28
  • I edited my answer accordingly. I didn't compute it explicitly, but i hope it helps. – ugur efem May 01 '16 at 11:31
  • Thank you :) how does the homogeneous condition change things? – thinker May 01 '16 at 12:20
  • homogeneous means that not just the polynomials but also all monomials must be of degree $d$. So for example if $d=3$ then $V$ does not contain $-x_1x_2x_3+x_1^2$. You still need to assume $V$ also contains $0$ though – ugur efem May 01 '16 at 12:39

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The monomials $x_1^{d_1}x_2^{d_2}x_3^{d_3}$ with $d_1+d_2+d_3=d$ form a basis of $V$. By the Stars and Bars Theorem there are ${d+2\choose2}$ such monomials. It follows that ${\rm dim}(V)={d+2\choose 2}$.

Christian Blatter
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A basis for $V$ will be the following set: $$\{x_1^ix_2^jx_3^k : i+j+k=d\}$$

So what you looking is essentially number of different partitions of $d$ using 3 natural numbers (called a restricted partition). I think the following wikipedia page includes the answer to that https://en.wikipedia.org/wiki/Partition_(number_theory)

ugur efem
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