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How would you find the determinant of an $m \times m$ matrix which has $m$ as every diagonal entry and $-1$ as every non diagonal entry?

darij grinberg
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Amanda
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  • Add $1/m$ times row $1$ to all other rows. Then use cofactor expansion along the first column. – apnorton Apr 24 '14 at 02:01
  • I believe the eigenvalues would be $m-1$ $(m+1)$'s and 1 $1$, so the product would be $(m+1)^{m-1}$. Can anyone verify this? –  Apr 24 '14 at 02:03
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    This might prove useful: http://math.stackexchange.com/questions/757320/prove-determinant-of-n-times-n-matrix-is-an-1ba-bn-1/757342#757342 – Brian Fitzpatrick Apr 24 '14 at 02:09

2 Answers2

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Let $A$ be the $m\times m$ matrix with $0$ on the diagonal and $-1$ elsewhere. The determinant you want can be found by evaluating $$\det(A-\lambda I)$$ and then substituting $\lambda=-m$. Now it is easy to see that if $\lambda=1$ then $A-\lambda I$ has a row echelon form with $m-1$ zero rows. Therefore $1$ is an eigenvalue of $A$ with geometric multiplicity $m-1$ and hence algebraic multiplicity at least $m-1$. The last eigenvalue $\mu$ is found from the trace, $$(m-1)(1)+\mu={\rm trace}(A)=0$$ and so $\mu=-(m-1)$. So $A$ has characteristic polynomial $$\det(A-\lambda I)=(-1)^m(\lambda-1)^{m-1}(\lambda+m-1)\ ,$$ and the determinant of your matrix is $$\det(A+mI)=(-1)^m(-m-1)^{m-1}(-m+m-1)=(m+1)^{m-1}\ .$$

David
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Your matrix is can be written as a rank $1$ perturbation to identity $$(m+1)I_m - \begin{bmatrix}1\\ 1\\ \vdots\\ 1\end{bmatrix}\begin{bmatrix}1& 1& \vdots& 1\end{bmatrix}$$ Now make use of Sylvester determinant theorem: $$\det(I+UV^T) = \det(I+V^TU)$$to obtain what you want after appropriate scaling.

user141421
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