7
$\begingroup$

Let, $V$ be a set of all homogeneous polynomial of degree $d$ in $n$-variables over a field $F$. Then dimension of $V_F$ is

(A) $\left(\begin{matrix}n\\d\end{matrix}\right)$

(B) $\left(\begin{matrix}d\\n\end{matrix}\right)$

(C) $\left(\begin{matrix}n+d-1\\d-1\end{matrix}\right)$

(D) $\left(\begin{matrix}n+d\\d\end{matrix}\right)$.

I tried through example for $n=2,3,4$ and $d=2,3,4$ and try to generalize the formula.

For $n=3$ and $d=3$ we get the basis of $V$ is $\{x^3,y^3,z^3,xy^2,yx^2,yz^2,zy^2,xz^2,zx^2,xyz\}$. So $dim(V)=10$. Similarly for $n=3$ and $d=2$ we get $dim(V)=6$. Also for $n=2$ and $d=2$ $dim(V)=3$. But I could not generalize these for arbitrary $n$ and $d$.

Suggest to find the general formula or any other way to determine it directly..

$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

The standard basis consists of all monomials $x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$, where the $a_i$ are non-negative integers such that $a_1+\cdots +a_n=d$.

By Stars and Bars the number of such $n$-tuples $(a_1,\dots,a_n)$ is $\binom{n+d-1}{n-1}$ or equivalently $\binom{n+d-1}{d}$.

There are also many questions/answers on MSE that use the Stars and Bars argument.

Remark: None of the suggested answers is correct!

$\endgroup$
5
  • $\begingroup$ Can you please give the link of wiki ? $\endgroup$
    – Empty
    Jun 1, 2015 at 3:19
  • $\begingroup$ OK, a minute or so. $\endgroup$
    – André Nicolas
    Jun 1, 2015 at 3:19
  • 1
    $\begingroup$ Sorry may be i am missing something but Is the set of all such polynomial a vector space? I mean sum of two polynomials of degree 3 say, need not be a polynomial of degree 3 know? Shouldn't it be polynomials of deg less equals d $\endgroup$
    – Meow
    Dec 6, 2016 at 4:37
  • $\begingroup$ @Prajakta If I sum two polynomials of degree 3 (say f = g + h), then f will also be a polynomial of degree 3 where the coefficients in each term of f are the sum of the coefficients of the corresponding terms in g and h. I believe you may be thinking of that fact that if we multiply two such polynomials, then we won't get back a polynomial of degree 3 (in fact, the degree will be 6). $\endgroup$
    – wanderingmathematician
    Jan 19, 2017 at 16:42
  • $\begingroup$ @wanderingmathematician I know this is old, but the original comment was correct in that the sum of polynomials of degree $n$ is of degree at most $n$ but not necessarily of degree $n$, since the leading coefficients can cancel in addition, e.g. if $f\left(x\right) = x + 1$, $g\left(x\right) = - x + 1$, in which case $f\left(x\right) + g \left(x\right) = 2$, hence $\deg \left(f + g\right) = 0 < 1 = \deg f = \deg g$. The same reasoning (plus consideration of the case where $f = 0$ and/or $g = 0$) gives the more general result that $$\deg \left(f + g\right) \leq \max \{\deg f, \deg g\}$$. $\endgroup$
    – Sam Streeter
    Nov 25, 2019 at 15:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .