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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correction made – cele Jul 01 '14 at 17:44
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And now the title is also fixed. – Hans Engler Jul 01 '14 at 17:48
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Did you try it with $n=1,2,3,4$? – Jonas Meyer Jul 01 '14 at 17:49
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This is a rank 1 perturbation of a diagonal matrix, so look at this post: http://math.stackexchange.com/questions/730134/determinant-of-rank-one-perturbation-of-a-diagonal-matrix – Hans Engler Jul 01 '14 at 17:50
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@JonasMeyer no, I did not. I wasnt sure what to do at first. – cele Jul 01 '14 at 23:30
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I suggest, when you don't know what to do at first, try something simple to get started. In this case, starting with $n=1$, then $n=2$, etc., may help. – Jonas Meyer Jul 01 '14 at 23:41
2 Answers
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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.
Rene Schipperus
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We posted very near answers at nearly the same time. I deleted mine. – Jonas Meyer Jul 01 '14 at 18:10
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I noticed that. There was no need to delete it, it would have been upvoted. – Rene Schipperus Jul 01 '14 at 18:12
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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! – Robert Lewis Jul 01 '14 at 18:52
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@RobertLewis, I agree. I have looked at different solutions to the same problem and found myself confused with some approaches but clearer with others. – cele Jul 01 '14 at 23:32