5

Let $F$ be a subfield of the complex numbers. We define $n$ linear functionals on $F^n$ ($n \geq 2$) by $f_k(x_1, \dotsc, x_n) = \sum_{j=1}^n (k-j) x_j$, $1 \leq k \leq n$. What is the dimension of the subspace annihilated by $f_1, \dotsc, f_n$?

Approaches I've tried so far:

  1. Construct a matrix $A$ whose $k$th row's entries are the coefficients of $f_k$, i.e., $A_{ij} = i - j$, and compute the rank of the matrix. Empirically, the resulting matrix has rank 2 for $n = 2$ to $n = 6$, but I don't see a convenient pattern to follow for a row reduction type proof for the general case.
  2. Observe that $f_k$, $k \geq 2$ annihilates $x = (x_1,\dotsc, x_n)$ iff $$\sum_{j=1}^{k-1} (k-j)x_j + \sum_{j = k+1}^n (j-k)x_j = 0$$ iff $$k\left(\sum_{j=1}^{k-1} x_j - \sum_{j=k+1}^n x_j\right) = \sum_{j=1}^{k-1} jx_j - \sum_{j=k+1}^n jx_j,$$ and go from there, but I do not see how to proceed.

Note: This is Exercise 10 in Section 3.5 ("Linear Functionals") in Linear Algebra by Hoffman and Kunze; eigenvalues and determinants have not yet been introduced and so I am looking for direction towards an elementary proof.

user169845
  • 745

3 Answers3

4

All the $f_k$ are linear combinations of the two linear functionals $$ \sum_{j=1}^n x_j \quad\text{and}\quad \sum_{j=1}^n jx_j; $$ therefore the dimension is at most $2$. Checking that the dimension is at least $2$ should be easy.

(For an exercise, you might want to use this observation to construct a solution along the lines of your idea #1.)

Greg Martin
  • 78,820
1

Here is the proof for rank of $A$ being 2. It is from @daw given as answer to another question with a little edition.

It is clear for $n=2$. Let $n\ge3$. And take $i\ge 3$. Then $$ a_{ij} - a_{1j} = i-1, \ a_{2j}-a_{1j} = 1, $$ which implies $$ a_{ij} - a_{1j}- (i-1)(a_{2j}-a_{1j}) =0. $$ So rank of $A$ is two, as the first two rows are clearly linearly independent.

Majid
  • 3,357
0

Dimension of the subspace annihilated by $f_i$'s is the dimension of the solution space of the system $$AX=0$$ where the $m^\text{th}$ row of $A$ is defined by $$A_m=(m-1,m-2,\dots,m-n),1\le m\le n.$$ For $m\ge 3,$ perform the following elementary operations \begin{align} &(1)A_m\to(1-m)A_2+A_m\\ &(2)A_m\to \frac{1}{(m-2)} A_m\\ &(3)A_m\to A_m+A_1 \end{align} On performing $(1)$ $$A_m=\big(0,m-2,2(m-2),3(m-2),\dots,(n-1)(m-2)\big).$$ On $(2)$ $$A_m=-A_1.$$ On $(3)$ $$A_m=0.$$ It is evident that $A_1$ and $A_2$ are linearly independent.

Thus, the dimension of the subspace annihilated = $n-2.$

Bijesh K.S
  • 2,604