# Tag Info

51

The formula you're looking for can be obtained by differentiating Jacobi's formula $$\frac{\mathrm{d}}{\mathrm{d}t} \det A(t) = \det A(t) \cdot \operatorname{tr}\left( A^{-1} \frac{\mathrm{d}A}{\mathrm{d}t} \right)$$ with respect to a second parameter, say $s$: \begin{multline} \frac{\partial^2}{\partial s \partial t} \det A(s,t) = \det A(s,t) \cdot \bigg[ ...

33

At first, we prove that the determinant is non-zero, in other words, the matrix is non-singular. Assume the contrary, then by the linear dependency of the columns there exist real numbers $\lambda_1,\dots,\lambda_n$, not all equal to 0, such that $F(x_i):=\sum_j \frac{\lambda_j}{g(x_i-b_j)}=0$ for all $i=1,2,\dots,n$. But the equation $F(x)=0$ is a ...

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Not all has been said about this question that is worth saying -- at the very least, someone could have written down the version without the absolute values; but more importantly, there are various other equally good proofs. Notations and statement Let me first state the result with proper signs and no absolute values. Standing assumptions. The following ...

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We can just manipulate $C$ in the usual way by row operations: Subtract the last "row" from all the other "rows" (this is really several traditional row operations done at once). This produces $$\begin{pmatrix} A- B &0& 0 & \ldots & 0 &B-A \\ 0 & A-B &0 &\ldots & 0 & B-A\\ && \ldots &&&\\ B & B ... 16 The answer is YES in every dimension, up to a sign. Here is the calculation. On the one hand, \det \tilde A=\det(A^2+I) because the blocs commute to each other. Therefore$${\rm Pf}(\tilde A)^2=\det(I+iA)\det(I-iA).$$On the other hand$$\det(I-iA)=\overline{\det(I+iA)}=\det(I+iA)^*=\det(I+iA),$$yields$${\rm Pf}(\tilde A)^2=(\det(I+iA))^2.$$Let us ... 16 Your guess is correct. If the elements outside the diagonal have absolute values less than 1/(n-1), the matrix has 'diagonal dominance', thus it is nonsingular. To make the answer self-contained, I give a proof. If x=(x_1,\dots,x_n)^t satisfies Ax=0, take k such that |x_k| is maximal and look at \sum a_{ki}x_i. The summand a_{kk}x_k has ... 15 let me assume A is invertible, then you ask when$$\det(1+X)=1+\det X,\;\;X=A^{-1}B $$so if X has eigenvalues x_i, i=1,2,\ldots n, you would need$$\prod_{i}(1+x_i)=1+\prod_i x_i$$basically you can take arbitrary values for x_1,x_2,\ldots x_{n-1} and then the only requirement is that$$x_n=\frac{1-U}{U-V},\;\;U=\prod_{i=1}^{n-1}(1+x_i),\;\;V=\...

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Darij already gave a great elementary detailed exposition. I wanted to remark that all such kinds of results (including the postscript) are special cases of a classical result that (at least in the generality of semilattices) goes back to Lindström in Determinants on semilattices. For this question we are in the special case of the Boolean lattice.

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Update 1: Thanks to jjcale for pointing out a fatal flaw. Indeed $SO(n,n)$ has two components, see here, and it looks suspiciously like my $T$ below is in the wrong component. I don't really know what Wikipedia means by "preserving/reversing orientation," but certainly $T=\textrm{diag}(-1,1,-1,1)$ is in the wrong component, and my $T$ below feels like it ...

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