TITLE: Is the matrix $\left({2m\choose 2j-i}\right)_{i,j=1}^{2m-1}$ nonsingular?
QUESTION [26 upvotes]: Suppose we have a $(2m-1) \times (2m-1)$ matrix defined as follows:
$$\left({2m\choose 2j-i}\right)_{i,j=1}^{2m-1}.$$
For example, if $m=3$, the matrix is 
$$\begin{pmatrix}6 & 20 & 6& 0 & 0\newline 1 & 15 & 15 & 1 & 0 \newline 0 & 6 & 20 & 6 & 0 \newline 0 & 1 & 15 & 15 & 1 \newline 0 & 0 & 6 & 20 & 6 \end{pmatrix}$$ 
Can anyone tell me how to prove it is nonsingular?

REPLY [21 votes]: Here is a very low-brow answer to the original question.
Consider the lower-triangular matrix
\begin{equation*}
V = [V_{ij}] = \left[\binom{i-1}{j-1}\right]\quad \text{for}\quad i \ge j.
\end{equation*}
Let $A$ be the matrix in the OP. Then, a quick induction shows that $V^{-1}AV$ is upper triangular (use that $V^{-1}$ has entries $(-1)^{i-j}V_{ij}$), with the diagonal: $[2^n,2^{n-1},\ldots,2]$, which not only establishes invertibility of $A$, but also yields its eigenvalues.