Trace of logarithm of anti-diagonal matrix
Yes, this is true: the trace of $\log M$ in your setting is the sum of logarithms of antidiagonal values, up to an integer multiple of $\pi i$.
Matrix logarithm are a bit complicated due to non-uniqueness, so it's easier to think of matrix exponential instead. One of the fundamental properties of the matrix exponential is $$\det (e^A) = e^{\operatorname{tr}A}$$ Here $e^A=M$ is your matrix and $A$ is (any) matrix logarithm of $M$. The determinant of $M$ is the product of antidiagonal entries up to the sign, the sign being $(-1)^{n(n-1)/2}$. And the trace of $A$ is a logarithm of that; recall that the logarithm is defined up to an even multiple of $\pi i$. Thus, $$ \operatorname{tr}A = \sum_{k=1}^n \log a_k \pmod {\pi i} $$
nGlacTOwnS
Updated on December 27, 2020Comments
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nGlacTOwnS almost 3 years
Is it true in general that the trace of log of an anti-diagonal matrix equals to the sum of log of anti diagonal elements? For a definite example, I have encountered a problem in which I need to compute Trace[Log[M(p,q)]], where M(p,q) is a some function of the form $$M(p,q) = \eta(p) \delta(p+q) $$ Is it correct that the value of the above expression if $$\int dq \: \eta(q) \: \delta (0)$$ up to some constant? I have checked this for a few finite dimensional matrices on Mathematica and found it to be true:
$Assumptions = And[a > 0, b > 0, c > 0, d > 0, e > 0, f > 0, g > 0, h > 0]; mat = {{0, 0, 0, 0, 0, 0, 0, a}, {0, 0, 0, 0, 0, 0, b, 0}, {0, 0, 0, 0, 0, c, 0, 0}, {0, 0, 0, 0, d, 0, 0, 0}, {0, 0, 0, e, 0, 0, 0, 0}, {0, 0, f, 0, 0, 0, 0, 0}, {0, g, 0, 0, 0, 0, 0, 0}, {h, 0, 0, 0, 0, 0, 0, 0}}; Tr[MatrixLog[mat]] // FullSimplify
The answer is
4 I \[Pi] + Log[a b c d e f g h]
(same was true for some higher dimensions too. The constant $4 i \pi$ is inconsequential for my purpose)
I was wondering if it is true in general and can it be proved to be so?