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Is it possible to quickly evaluate the following matrix is rank 1 or not? \begin{pmatrix} 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 \\ -1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 \\ -1 & -1 & 1 & -1 & -1 & 1 & 1 & 1 \\ 1 & -1 & -1 & 1 & -1 & -1 & 1 & 1 \\ 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 \\ 1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 \\ -1 & 1 & 1 & 1 & -1 & -1 & 1 & -1 \\ -1 & -1 & 1 & 1 & 1 & -1 & -1 & 1 \end{pmatrix}

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2 Answers 2

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Similarly, if a matrix is of rank $1$, then its column/row space has dimension $1$, namely if you fix a column/row, all other columns/rows are scalar multiples of the fixed one.

Apparently, your new matrix is also not of rank $1$.

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If a matrix is rank $1$, then each of its $2\times 2$ submatrices has zero determinant. However, the top left $2\times 2$ submatrix of your matrix has determinant $-1$. So, your matrix is not rank $1$.

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  • $\begingroup$ sorry, I mistakenly wrote the diagonal to be zero, and in fact it is 1. And, why "If a matrix is rank 1, then each of its 2×2 submatrices has zero determinant"? Could you provide some references? $\endgroup$
    – M.K
    Jan 7 at 21:47
  • $\begingroup$ @M.K It is about principal minors. Check this: en.wikipedia.org/wiki/Minor_(linear_algebra)#Other_applications $\endgroup$ Jan 7 at 21:50

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