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This question already has an answer here:

Let $n$ be a positive integer and let $A=[a_{ij}] \in M_{n\times n} (R)$ be the matrix defined by

$a_{ij}=0$ if $i=j$
$1$ otherwise

To be honest, I've only calculated determinants of matrices with numbers, nothing like this.

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marked as duplicate by Martin Sleziak, user147263, user1551 linear-algebra Jul 1 '14 at 20:19

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.

  • $\begingroup$ correction made $\endgroup$ – cele Jul 1 '14 at 17:44
  • $\begingroup$ And now the title is also fixed. $\endgroup$ – Hans Engler Jul 1 '14 at 17:48
  • $\begingroup$ Did you try it with $n=1,2,3,4$? $\endgroup$ – Jonas Meyer Jul 1 '14 at 17:49
  • $\begingroup$ This is a rank 1 perturbation of a diagonal matrix, so look at this post: math.stackexchange.com/questions/730134/… $\endgroup$ – Hans Engler Jul 1 '14 at 17:50
  • $\begingroup$ @JonasMeyer no, I did not. I wasnt sure what to do at first. $\endgroup$ – cele Jul 1 '14 at 23:30
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Add all the columns to the first one and then subtract the first row from the others we find

$$\left|\begin{array}\\ n-1&1&\cdots&1\\ 0&-1&\cdots&0\\ \vdots&\cdots&-1&0\\ 0&\cdots&&-1 \end{array}\right|=(-1)^{n-1}(n-1)$$

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Another way is to use the fact that the determinant is the product of the eigenvalues. Let $B$ be the matrix with ones everywhere. Then $A=B-I$ it is easy to see that the eigenvalues of $B$ are $n$ with multiplicity $1$ and $0$ with multiplicity $n-1$. Now if $(B-I)v=\lambda v$ then $Bv=(\lambda +1)v$ so the eigenvalues of $B$ are $n-1$ with multiplicity $1$ and $-1$ with multiplicity $n-1$. This gives $(-1)^{n-1}(n-1)$ in agreement with the answer of Sami.

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  • $\begingroup$ We posted very near answers at nearly the same time. I deleted mine. $\endgroup$ – Jonas Meyer Jul 1 '14 at 18:10
  • $\begingroup$ I noticed that. There was no need to delete it, it would have been upvoted. $\endgroup$ – Rene Schipperus Jul 1 '14 at 18:12
  • $\begingroup$ I agree it is not needed, but I think it is more useful deleted. $\endgroup$ – Jonas Meyer Jul 1 '14 at 18:13
  • $\begingroup$ I think that often times different verbalizations of the same result, proof, solution, etc., will appeal to, or be clear to, different readers. So I don't mind multiple expressions of the same line of thought. You never know what will "hit" a given reader. Cheers! $\endgroup$ – Robert Lewis Jul 1 '14 at 18:52
  • $\begingroup$ @RobertLewis, I agree. I have looked at different solutions to the same problem and found myself confused with some approaches but clearer with others. $\endgroup$ – cele Jul 1 '14 at 23:32

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