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I need to calculate the determinant of the following matrix:$$\begin{bmatrix}0&0&-2x_1& \cdots &-2x_n&0& \cdots &0\\0&0&0& \cdots&0&-2x_1& \cdots&-2x_n\\-2x_1&0&-1& \cdots&0&1& \cdots&0\\ \vdots&\vdots &\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\-2x_n&0&0&\cdots&-1&0&\cdots&1\\0&-2x_1&1&\cdots&0&-1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\0&-2x_n&0&\cdots&1&0&\cdots&-1 \end{bmatrix}$$

Consider that all the elements that are only on the diagonals denoted by the dots are $1$ or $-1$.

Let $H_j$ be the upper leftmost $j$ x $j$ submatrix of A. I need a method to evaluate all the determinants for which $5\le j\le2n+2$. Could someone give me an advice?

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  • $\begingroup$ You may write $\vec x$ $\endgroup$ – Dude Nov 6 '14 at 17:43
  • $\begingroup$ How did you get it? $\endgroup$ – Dude Nov 6 '14 at 18:37
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Write w.l.o.g. your matrix in the form $$ A:=\begin{bmatrix} 0 & 0 & x^T & 0 \\ 0 & 0 & 0 & x^T \\ x & 0 & -I & I \\ 0 & x & I & -I \end{bmatrix}. $$ Take any nonzero vector $z$ orthogonal to $x$ ($z^Tx=0$), there is a whole $(n-1)$-dimensional subspace of them. Now $$ A\begin{bmatrix} 0 \\ 0 \\ z \\ z \end{bmatrix}=0. $$ Hence for $n>1$, $A$ is singular and $\det(A)=0$.

For $n=1$, it is easy to verify that $\det(A)=x^4$.

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