Kernel of a specific antisymmetric matrix I am trying to compute the kernel of the following real antisymmetric $m \times m$-matrix:
$A = (a_{ij})$, where $a_{ij} =  \begin{cases} 0, & i = j, \\
      \lambda_{j-i}, & j > i, \\
      -\lambda_{i-j}, & j < i.
             \end{cases}$
Moreover the following data are known


*

*$m = 2k+1$ is odd

*all $\lambda_i$ are non-zero

*$\lambda_i = -\lambda_{m-i}$ for all $i$ (in particular, $A$ is determined by $\lambda_1, ..., \lambda_{\lfloor m/2 \rfloor}$)


I am pretty sure that the kernel is generated by $(1,...,1)^T$, but I cannot prove it. Here's what I've done so far:


*

*Computation of examples via software

*Try to see the kernel via elementary row operations

*Induction on $k$ to show that the rank of $A$ is $2k$ (then one would be done, because $(1,...,1)^T$ obviously lies in the kernel), in the following way: setting $\lambda_1 = 1$, $A$ has the form $\begin{pmatrix} C & B \\ -B^T & A' \end{pmatrix}$, where $C = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$. Using elementary row operations, one can see remodel $A$ to the matrix $\begin{pmatrix} C & B \\ 0 & A' + B^TC^{-1}B \end{pmatrix}$. The hope was now that $F := A' + B^TC^{-1}B$ is of the same form as $A$, but with less parameters. But it seems like this is not the case. 
For instance, I have computed the case where $m=7$ in Maple. Here's a screenshot of $F$ in this case (for convenience, the $\lambda_i$ are called $ai$ in Maple).



If anyone knows how to compute the kernel in a good way, please give me a hint. I want to do the details of my own.

Your matrix is a Toeplitz matrix, which is amenable to LU-decomposition, from which you should be able to read off generators of the kernel. To do the LU decomposition, I suggest you reorder the rows, replacing row $i$ by row $n - i$. In the $3 \times 3$ case, that ends up looking like this:
\begin{align}
\begin{bmatrix}
-b & -a & 0 \\
-a & 0 & a \\
0 & a & b
\end{bmatrix} =
\begin{bmatrix}
1 & 0 & 0 \\
a/b & 1 & 0 \\
0 & b/a & 1
\end{bmatrix} \cdot
\begin{bmatrix}
-b & -a & 0 \\
0 & a^2/b & a \\
0 & 0 & 0
\end{bmatrix} \cdot
\end{align}
Clearly in this case the diagonals of the $U$ matrix show that there's a 1-dimensional kernel. 
My guess is that if you work out the $2 \times 2$ and $4 \times 4$ cases, you'll see a pattern and be done. 
I also suspect that since you've got a good guess of the kernel-generator (all 1s), you could probably rewrite this matrix in a different basis consisting of this vector (call it $b_n$) and a bunch of other vectors that are orthogonal to it, such as those of the form $e_k - e_1$ for $k = 2, \ldots n$. The change of basis matrix in this case may be invertible "by eye", in which case it may be obvious that the upper left $(n-1) \times (n-1)$ submatrix of the conjugate of $A$ by this change-of-basis is invertible. But that's just a conjecture. 
