I am trying to solve the following exercise:
Problem
Let $F$ be a subfield of the complex numbers. We define $n$ linear functionals in $F^n$ ($n \geq 2$) with $$f_k(x_1,…,x_n)=\sum_{j=1}^n(k-j)x_j$$
What is the dimension of the subspace annihilated by $f_1,…,f_n$?
I am not so sure what to do here. By the way these functionals are defined, it is clear that the $k-th$ coordinate of any vector $\alpha$ in $F^n$ is "annihilated" by the functional $f_k$. For example, if I consider the standard basis of the vector space, $\mathcal B=\{e_1,…,e_n\}$, then $f_i(e_j)=\delta_{ij}$. I don't realize how to deduce from these facts the dimension of the subspace. Any help would be appreciated.
Best Answer
You can try thinking of the linear transformation $A: F^n \rightarrow F^n$ which is made up of the $n$ functionals, i.e.,
$Ax = (f_1(x),\ldots,f_n(x))$
and so the question becomes finding the dimension of the kernel. With respect to the standard basis, in the special case $n=4$, $A$ has matrix $$ \left(\begin{array}{rrrr} 0 & -1 & -2 & -3 \\ 1 & 0 & -1 & -2 \\ 2 & 1 & 0 & -1 \\ 3 & 2 & 1 & 0 \end{array}\right) $$
Row reduce using the second row to see what the rank is. Your answer for the rank will hold for all $n$, not just 4.