= Solution
The <monomials> $x^iy^j$ with $i,j\ge0$ and $i+j\le n$ form a <basis> of the given <vector space>. There are $k+1$ of <total degree> $k$, so
$$
\boxed{\dim V=\sum_{k=0}^n(k+1)=\frac{(n+1)(n+2)}2.}
$$
On that <basis> the <linear operator> is diagonal:
$$
S(x^iy^j)=\bigl(i(i-1)+j(j-1)\bigr)x^iy^j.
$$
Both summands of the <eigenvalue> are nonnegative integers. Their sum is zero precisely when $i,j\in\{0,1\}$. Thus \b[the <kernel> is the span of those members of $1,x,y,xy$ whose <total degree> is at most $n$]. In particular it is $\operatorname{span}\{1,x,y\}$ for $n=1$ and $\operatorname{span}\{1,x,y,xy\}$ for $n\ge2$; for $n=0$ it is the constants. \b[The <image of a linear map> is the span of all the other permitted <monomials>], since each has a nonzero diagonal <eigenvalue> and hence is the image of its reciprocal multiple. No excluded <monomial> can occur in an image because its diagonal coefficient is zero.
The <trace> is the sum of these <eigenvalues>. At degrees zero and one all contributions vanish. At degree two the contributions are $2,0,2$, summing to four; at degree three they are $6,2,2,6$, summing to sixteen; at degree four they are $12,6,4,6,12$, summing to forty. Therefore
$$
\boxed{\operatorname{tr}S=0,4,20,60\quad\text{for }n=1,2,3,4\text{ respectively}.}
$$
This is the <diagonal monomial action of a second-order Euler operator>.
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