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's non-singular?

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?

Suppose we have a (2m-1) by (2m-1) matrix as following: $$(C_{2j-i}^{2m})_{i,j=1}^{2m-1}.$$ Here $C_p^q$ is defined to be $\frac{q!}{p!(q-p)!}.$ Can any one tell me how to prove it's non singular?

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's non-singular?