Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
I am interested in why you would want to find the formula for such an expression. –  fpqc Sep 4 '12 at 7:40
1  
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
add comment

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!)

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.