Prove that this determinant equals zero

Question

I recently came up with a problem in which I need to use the fact that the determinant below is equal to zero :

\begin{vmatrix} 0 & a_{21} & 0 & a_{41} & 0 & \dots&0 &a_{2k-2,1}&0&a_{2k,1}&0\\ a_{12} & 0 & a_{32} & 0&a_{52}& \dots & a_{2k-3,2}& 0& a_{2k-1,2} & 0 & a_{2k+1,2} \\ \vdots & \vdots & \vdots &\vdots& \vdots & \dots&\vdots&\vdots&\vdots &\vdots&\vdots \\ a_{1,2k} & 0 & a_{3,2k} & 0&a_{5,2k}& \dots &a_{2k-3,2k}&0&a_{2k-1,2k}&0&a_{2k+1,2k} \\ 0&a_{2,2k+1}&0&a_{4,2k+1}&0&\dots & 0&a_{2k-2,2k+1}&0&a_{2k,2k+1}&0 \end{vmatrix}

So it means that the elements $a_{ij}=0$ for $i+j=2n$ where $n$ is a natural number

I don't know how to prove that the above determinant is zero , using elementary transformations or using the induction hypothesis it gets very complicated , so is there any easy way to prove it, thanks a lot in advance .

Answer 1

Hint: Maybe you can consider the definition of determinant:

In the $1,3,5,\cdots,(2k+1)$ rows(in total $(k+1)$ rows), the nonzero elements are in the $2,4,\cdots,2k$ columns(in total $k$ columns), so according to the drawer principle, you will always get $0$ in the definition of determinant.

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The formula named Leibniz formula in Wikipedia is the definition of $n\times n$ determinant in my book:

Answer 2

I'll express it in a slightly different way: there are $k+1$ rows with an odd index, and each depends on $k$ coefficients (the same coordinates). So they correspond to a set of $k+1$ vectors in $F^k$ ($F$ denoting the base field). $k+1$ vectors in a space of dimension $k$ are linearly dependent. Hence the odd-numbered rows are linearly dependent, and you can replace any of them, through elementary operations on rows, with a null row.