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Consider $A_{n\times n}$ is an asymmetric matrix over a field. We denote the transpose of matrix $A$ with $A^T$. Suppose that the matrix $A$ has this property that all it's sub-matrices of order $i$, $1\leq i \leq n$, have non-zero determinant. We can see that because of $A$ has this property then all sub-matrices $A^T$ of order $i$, $1\leq i \leq n$ have non-zero determinant too.

Assume matrix $B_{n \times 2n}$ is constructed by the matrices $A$ and $A^T$, in the following form $$ B=[A|A^T] $$ My question: Is it true claim that all sub-matrices of order $j$, $1\leq j \leq n$, of the matrix $B$ have non-zero determinant?

We know that for a sub-matrix of $B$ of order $i$, we can choose $i$ number of set $[1..n]$ for rows ,$n \choose i$, and $i$ number of set $[1..2n]$ for choosing columns, $2n \choose i$.

For example: consider the following form of the companion matrix ($n>2$) $$ C_n=\left( \begin{array}{cccccc} 0 &1 &0 &\cdots &\cdots &0 \\ 0 &0 &1 &\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots &\ddots &\ddots &0 \\ 0 &\cdots &\cdots &0&0 &1 \\ 1 &1 &1 &\cdots &1 &1 \end{array} \right)_{n\times n} $$ with numerical computation, I have seen that the $n\times (n-1)$ power of matrix $C_n$ has this property and all sub-matrices of matrix $B=[C_n^{n^2-n}|{(C_n^{n^2-n})}^T]$ have non-zero determinant.

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No.

Take $A = \begin{pmatrix}1 & 2 \\ 1 & 4\end{pmatrix}$

Then $B = \begin{pmatrix}1 & 2 & 1 & 1 \\ 1 & 4 & 2 & 4\end{pmatrix}$.

Now take the second and third column to get a determinant of zero. :)

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  • $\begingroup$ Thanks, you mean second and third columns! $\endgroup$ – Amin235 Apr 11 '17 at 14:08

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