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This question already has an answer here:

Can someone please explain how to compute the determinant of $J_n - I_n$ where $j_n$ it a matrix of ones?

E.g. for $n=5$ we get the following matrix

$$\left(\begin{array}{ccccc} 0 & 1 & 1 &1 &1 \\ 1 & 0 & 1 &1 &1 \\ 1 & 1 & 0 &1 &1 \\ 1 & 1 & 1 &0 &1 \\ 1 & 1 & 1 &1 &0 \end{array} \right)$$

By computing it for the first few $n$'s it seems to be $(-1)^{n+1}(n-1)$ but I couldn't prove it.

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marked as duplicate by Marc van Leeuwen, user1551, DonAntonio, azimut, Antonio Vargas Mar 30 '13 at 14:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let us put

$$ D =\left|\begin{array}{ccccc} 0 & 1 & 1 &1 &1 \\ 1 & 0 & 1 &1 &1 \\ 1 & 1 & 0 &1 &1 \\ 1 & 1 & 1 &0 &1 \\ 1 & 1 & 1 &1 &0 \end{array} \right| $$

Performing the operation $L_1 \gets \sum_{k=1} L_k$ and putting $m=n-1$, we obtain

$$ D =\left|\begin{array}{ccccc} m & m & m &m &m \\ 1 & 0 & 1 &1 &1 \\ 1 & 1 & 0 &1 &1 \\ 1 & 1 & 1 &0 &1 \\ 1 & 1 & 1 &1 &0 \end{array} \right|= m\left|\begin{array}{ccccc} 1 & 1 & 1 &1 &1 \\ 1 & 0 & 1 &1 &1 \\ 1 & 1 & 0 &1 &1 \\ 1 & 1 & 1 &0 &1 \\ 1 & 1 & 1 &1 &0 \end{array} \right| $$

Performing $L_j \gets L_j-L_1$ for all $j\gt 1$, we obtain

$$ D = m\left|\begin{array}{ccccc} 1 & 1 & 1 &1 &1 \\ 0 & -1 & 0 &0 &0 \\ 0 & 0 & -1 &0 &0 \\ 0 & 0 & 0 &-1 &0 \\ 0 & 0 & 0 &0 &-1 \\ \end{array} \right| =m(-1)^m $$

as you expected.

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  • $\begingroup$ +1 Eventhough this is a repeated question, this answer stands up by its clearity and simplicity. $\endgroup$ – DonAntonio Mar 30 '13 at 14:29
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Consider that $J_n=(1,\cdots,1)^T(1,\cdots,1)$ and robjohn's comment in the following: How to show if $ \lambda$ is an eigenvalue of $AB^{-1}$, then $ \lambda$ is an eigenvalue of $ B^{-1}A$?

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  • $\begingroup$ Blatantly missed it, robjohn's comment can be replaced by en.wikipedia.org/wiki/Sylvester%27s_determinant_theorem $\endgroup$ – Guest Mar 30 '13 at 14:24
  • $\begingroup$ @Guest, thank you for pointing out this. This formula is very handy for calculating spectra. $\endgroup$ – Easy Mar 30 '13 at 14:28
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Subtract the first row from the rows below it, then add row 2 through $n$ to the first row, to get a triangular matrix.

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Row addition doesn't alter the determinant. Subtracting the first row from the others, then adding each other row to the first, you'll get a lower triangular matrix, whose determinant is simply the product of its diagonal entries.

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