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Given $$M := \mbox{diag} (1, 2, \dots, n) - n \, I_n + n \,1_n 1_n^T$$compute the determinant of $M$.

Rodrigo de Azevedo
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Jasmine
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1 Answers1

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Hint: This is a rank one update of a diagonal matrix. In particular, we have $$ M = \pmatrix{ 1-n\\ &2-n\\ && \ddots\\ &&& 0 } + n \cdot xx^T $$ where $x = (1,\dots,1)^T$. We can find the determinant of this matrix using Sylvester's determinant theorem (more specifically, using the matrix determinant lemma).

Full Solution: Let $M(t) = M + tI$. Let $$ D(t) = \pmatrix{ 1-n+t\\ &2-n+t\\ && \ddots\\ &&& t } $$ We note that for $t \neq 0$ (and such that $D(t)$ is invertible), $$ \det[M(t)] = \det[D(t) + nxx^T] = \det[D(t)]\det[I + n[D(t)]^{-1}xx^T] = \\ \det[D(t)](1 + n\,x^T[D(t)]^{-1}x) = \\ \left( \prod_{k=1}^n (k-n+t)\right) \left(1 + n \sum_{k=1}^n\frac{1}{k - n + t} \right) = \\ \left( \prod_{k=1}^{n-1} (k-n+t)\right) \left(t\left[1 + n \sum_{k=1}^{n-1}\frac{1}{k - n + t} \right] + n\right) $$ We then have $$ \det(M) = \lim_{t \to 0} \det(M(t)) = \left( \prod_{k=1}^{n-1} (k-n)\right)\cdot n = (-1)^{n-1}\,n! $$

Ben Grossmann
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  • In retrospect, there's a much quicker (or equally quick) approach using Gaussian elimination. – Ben Grossmann Jul 21 '15 at 23:24
  • If $$M := \underbrace{\mbox{diag} (1, 2, \dots, n)}_{=:D} - n , I_n + n ,1_n 1_n^T$$ then $$\det(M) = \det (D - n I_n) + n 1_n^T \mbox{adj} (D_n - n I_n) 1_n = 0 + n (-1)^{n-1} (n-1)! = (-1)^{n-1} n!$$where we used the adjugate of a diagonal matrix, which is easy to compute. In this case, only the southeast corner of the adjugate is nonzero. – Rodrigo de Azevedo Jul 17 '16 at 00:32
  • I didn't know about the adjugate formula, but that definitely works – Ben Grossmann Jul 17 '16 at 02:37