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What can one say about the function $(t,A,B) \mapsto \det(tA + (1-t)B)$, with $t \in [0,1]$, $A$, $B$ square matrices, in my case, say, permutational matrices? Where such a function shows up? Hoping for some references.

user1551
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Nikolai Mnev
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    I am not sure that this is your case, but still if $A,B$ are symmetric and positive definite matrices then the function $t\mapsto \log\det(tA+(1-t)B)$ is concave on the segment $[0,1]$. In other words the function $A\mapsto \log\det A$ is concave on the cone of positive definite matrices. This is used in the theory of Monge-Ampere equations. – MKO Jul 17 '16 at 08:46
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    If $B$ is invertible, then $\mathrm{det}(tA+(1-t)B)=\mathrm{det}(B)\mathrm{det}(tAB^{-1}+(1-t)I)$ which can be expressed in terms of the characteristic polynomial for $AB^{-1}$. – Oscar Cunningham Jul 17 '16 at 08:50
  • thank you sva and Oscar. Both comments are very useful. – Nikolai Mnev Jul 17 '16 at 09:21
  • If permutational matrices we can interpret the matrix $tA+(1-t)B$ as a continuum of markov chains. In the extreme cases of $t=0$ or $t=1$ they will be deterministic ( 100% probability for each non-zero probability ) and we will be sure of row and column sum equal to 1 will be preserved for all other t in the interval. – mathreadler Jul 17 '16 at 09:24
  • Thank you mathreadler. Indeed in permutational case determinant of stochastic matrix is some wonderful function. Here is a discussion. http://math.stackexchange.com/questions/932548/interpretation-for-the-determinant-of-a-stochastic-matrix – Nikolai Mnev Jul 17 '16 at 10:08
  • If we rewrite to have symmetry $(t-1)A + (t+1)B$, $t\in[-1,1]$ and plot the determinant it gets suspiciously close to some monomials $t^n$. I don't quite see the connection between $n$ and matrix side, but seems to be related. – mathreadler Jul 17 '16 at 15:24

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