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The question, with no simplifications or motivation:

Let $A$ and $B$ be square matrices of the same size (with real or complex coefficients). What is the most reasonable formula one can find for the determinant $$\det((1-t)A + tB)$$ as a function of $t \in [0,1]$? If no reasonable formula exists, what can we say about these determinants?

So we're taking a line between two matrices $A$ and $B$, and computing the determinant along this line. When $A$ and $B$ are diagonal, say $$A = \operatorname{diag}(a_1,\ldots,a_n), B = \operatorname{diag}(b_1,\ldots,b_n),$$ then we can compute this directly: $$\begin{aligned} \det((1-t)A + tB) &= \det \operatorname{diag}((1-t)a_1 + tb_1, \ldots, (1-t)a_n + tb_n) \\ &= \prod_{j=1}^n ((1-t)a_j + tb_j). \end{aligned}$$ I'm not sure if this can be further simplified, but I'm sure someone can push things at least a tiny bit further than I have.

I'm most curious about the case where $A = I$ and each $(1-t)A + tB$ is assumed to be invertible. Here's what I know in this case: writing $$D(t) = \det((1-t)I - tB),$$ we can compute that $$ \dot{D}(t) = D(t) c(t)$$ where $$c(t) := \operatorname{trace}(((1-t)I + tB)^{-1}(B-I))$$ (a warning: I am not 100% sure this formula holds). Thus we can write $$D(t) = \exp\left(\int_0^t c(\tau) \; d\tau\right)$$ since $D(0) = 1$. I have no idea how to deal with the function $c(\tau)$ though. Any tips?

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I am interested in why you would want to find the formula for such an expression. –  fpqc Sep 4 '12 at 7:40
In the diagonal case where $A = I$ (which is the only case I personally need), I'm linearly interpolating between the unit cube in $\mathbb{R}^n$ and the $n$-box $[0,b_1]\times \cdots \times [0,b_n]$, and $D(t)$ gives me the signed volume of the box obtained at time $t$. The general case is pure curiosity induced by looking at the (easy) diagonal case, but I suppose it could be interpreted as a signed volume calculation for a more general linear interpolation between $n$-boxes. –  Alex Amenta Sep 4 '12 at 7:56
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1 Answer

I think I have an answer to the last case I mentioned ($A=I$, all $(1-t)I + tB$ invertible). The key is to write $$\begin{aligned} \int_0^t c(\tau) \; d\tau &= \operatorname{trace} \int_0^t ((1-\tau)I + \tau B)^{-1} (B-I) \; d\tau \\ &= \operatorname{trace} \log ((1-t)I + tB) \end{aligned} $$ using that $\frac{d}{dt}((1-t)I + tB) = B-I$ and $\frac{d}{dt}\log(A(t)) = A(t)^{-1} \frac{d}{dt} A'(t)$ (I think this is true!). Taking matrix logarithms here should be ok, since everything in sight is invertible. Then $$ D(t) = e^{\operatorname{trace} \log((1-t)I + tB)}. $$ Of course, this leaves one with the problem of computing a matrix logarithm, but in the case where $B$ is diagonal, this reduces to the first formula in my question. This could probably be generalised to the case of general invertible $A$ by computing $\dot{D}(t)$ as before. I'll do this tomorrow and make an edit with the results (or you can do it!)

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Ah, this is what happens when I learn matrix calculus as I go along - apparently $\det(A) = \exp(\operatorname{trace}(\log A)))$ is a standard formula! I'll leave this up as a lesson in self-discovery –  Alex Amenta Sep 4 '12 at 12:51
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