Prove a symmetric matrix has full rank

Question

I'm trying to prove this statement:

Let function $f(x) = c_1 \lvert x-1 \rvert + ...... + c_n \lvert x-n \rvert$. If $f(1)=f(2)=......=f(n)=0$, then $c_1=c_2=......=c_n=0$.

To solve $c_1,......,c_n$, I'm thinking this is equivalent to solving

\begin{array}{@{}cccc|c@{}} 0 & 1 & \cdots & n-1 & 0 \\ 1 & 0 & \cdots & n-2 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ n-1 & n-2 & ... & 0 & 0 \end{array}

And if I can show that the nxn matrix on the left has rank n, then I can use the rank-nullity theorem and prove the above claim. However, I couldn't find an easier way to do this; all I can observe is that this matrix is symmetric and it's not helping. Am I on the right track or is there any other way to show this?

Answer 1

Let $M_n$ be the relevant matrix. We show that $\det M_n=(-1)^{n-1}(n-1)2^{n-2}$.

Replace the first column with the first column, minus twice the second, plus the third, to get that the determinant of $M_n$ is the determinant of the same matrix but with the first column replaced by $[0,2,0,\dots,0]$. Now do the same with the first row (replace it with the first row, minus twice the second, plus the third). This shows that $$\det M_n=\begin{vmatrix}-4&2e_1^\intercal \\2e_1&M_{n-1}\end{vmatrix},$$ where $e_1=[1,0,\dots,0]$. Expanding by minors along the first column (and then along the first row) gives $$\det M_n=-4\det M_{n-1}-2\begin{vmatrix}2&0\\b&M_{n-2}\end{vmatrix}=-4\det M_{n-1}-4\det M_{n-2}$$ for some vector $b$. Now, given the base cases of $\det M_1=0$ and $\det M_2=-1$, we can prove $\det M_n=(-1)^{n-1}(n-1)2^{n-2}$ by induction on $n$.

Now, since $(-1)^{n-1}(n-1)2^{n-2}\neq 0$ for $n>1$, $M_n$ is non-singular, as desired.

Answer 2

You need to assume $n>1$. Here is another way.

Note that for $x \notin \{1,...,n\}$ $f$ is differentiable and for $x \in (k,k+1)$, with $k=1,...,n-1$ we have $f'(x) = -(c_1+\cdots + c_k) + (c_{k+1}+\cdots + c_n)$. Since $f$ attains an extreme in $(k,k+1)$ we must have $-(c_1+\cdots + c_k) + (c_{k+1}+\cdots + c_n) = 0$. By subtracting the $k+1$ equation from the $k$ equation we get $c_2 = c_3 = \cdots = c_n = 0$. It then follows from $f(2) = 0$ that $c_1 = 0$.