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For a given function

$$h(A) = \lim\limits_{n \to +\infty} \dfrac{\text{tr}((A^A)^{n+1})}{\text{tr}((A^A)^{n})}$$

we have to check if it's a valid matrix norm.

I know that $A^A$ is defined as

$$A^A = \exp(A\ln(A))$$

where both $\exp(X)$ and $\ln(X)$ defined as infinite sum of series.

It's completely unclear though, how this definition may be of help to solve the problem. Not sure where to start and how to make any progress on this one. Maybe anyone could give an idea to the solution?

Martin Sleziak
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Swistack
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  • I might try the one-dimensional case where $A:\mathbb R\to\mathbb R$ is simply the multiply-by-constant operator, for which we can identify tr$(A)=A$. – user254433 Sep 18 '18 at 21:52

1 Answers1

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Let $\lambda_1, \ldots, \lambda_n$ be the eigenvalues of $A$ with multiplicities. The spectral mapping theorem implies that $\lambda_1^{\lambda_1}, \ldots, \lambda_n^{\lambda_n}$ are the eigenvalues of $A^A$, with the same multiplicities. Denote $x_i = \lambda_i^{\lambda_i}$.

WLOG assume that $|x_1| \le |x_2| \le \cdots \le |x_n|$.

\begin{align}\lim_{k\to\infty} \frac{\operatorname{Tr} (A^A)^{k+1}}{\operatorname{Tr} (A^A)^{k}} &= \lim_{k\to\infty} \frac{x_1^{k+1} + x_2^{k+1} \ +\cdots + x_n^{k+1}}{x_1^{k} + x_2^{k} \ +\cdots + x_n^{k}}\\ &= x_n \lim_{k\to\infty}\frac{\left(\frac{x_1}{x_n}\right)^{k+1} + \left(\frac{x_2}{x_n}\right)^{k+1} +\cdots + \left(\frac{x_{n-1}}{x_n}\right)^{k+1} + 1}{\left(\frac{x_1}{x_n}\right)^{k} + \left(\frac{x_2}{x_n}\right)^{k} +\cdots + \left(\frac{x_{n-1}}{x_n}\right)^{k} + 1}\\ &= x_n \end{align}

This is not a norm. For the identity matrix we have $h(I) = 1^1 = 1$ but $h(2I) = 2^2= 4$.

mechanodroid
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  • Thanks for the answer! I checked spectral mapping theorem, as far as I know this theorem applies only to polynomials e.g. if f is a polynomial and $\lambda$ is eigenvalue of $A$, then $f(\lambda)$ is eigenvalue of $f(A)$. But can we really make any such claims about spectrum of $A^A$? – Swistack Sep 18 '18 at 23:07
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    @Swistack Actually, the spectral mapping theorem is valid for functions which are holomorphic on some open neighbourhood of $\sigma(A)$. Have a look here for a statement in much greater generality. The claim about the equality of multiplicities follows from the Jordan form. – mechanodroid Sep 19 '18 at 00:20