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Suppose $v \in \mathbb R^n$ is a fixed vector. We define a scalar-valued function on $n \times n$ matrices $f: M_n(\mathbb R) \to \mathbb R$ by \begin{align*} A \mapsto \sum\limits_{j=0}^{\infty} \langle A^j v, A^j v \rangle. \end{align*} Let us denote the domain of $f$ by $\text{Dom}(f) = \{A \in M_n(\mathbb R): f(A) < \infty\}$.

It is clear if $\rho(A) < 1$ (spectral radius), then $A \in \text{Dom}(f)$. If I am not mistaken, $f$ should also be differentiable on $\{A: \rho(A) < 1\}$. On the other hand, if $\rho(A) \ge 1$, it is still possible $A \in \text{Dom}(f)$. For example, if $v$ is chosen to be an eigenvector corresponding to an eigenvalue strictly smaller than $1$.

The question bugs me is: could the function be differentiable on the set $\text{Dom(f)} \setminus \{A:\rho(A) < 1\}$.

MyCindy2012
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I believe $\{A:\rho(A)<1\}$ is the interior of $\mbox{Dom}(f)$, which means the answer is no. Since the inclusion "$\subseteq$" is pretty simple, I will only argue that for all $A\in\mbox{Dom}(f)$ with $\rho(A)\geq1$ there exists $B\not\in\mbox{Dom}(f)$ with $\|A-B\|$ arbitrarily small.

Let $A\in\mbox{Dom}(f)$ with $\rho(A)\geq1$. By changing $A$ an arbitrarily small amount we can obtain a matrix $B$ with a complex eigenbasis $\beta$, such that $\rho(B)\geq1$, and such that the representation of $v$ in terms of $\beta$ has only non-zero coordinates. Then for some eigenvector $b$ of $B$ the corresponding eigenvalue is at least $1$ in absolute value. Then the absolute value of the $b$-coordinate of $B^jv$ with respect to $\beta$ is non-decreasing. It follows that $\langle B^jv,B^jv\rangle$ does not converge to $0$, so the series $f(B)$ does not converge, so $B\not\in\mbox{Dom}(f)$.

SmileyCraft
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  • Thanks for answering the question. Can the function be differentiable in a restricted sense: for any $A \in \text{Dom}(f) \setminus {A: \rho(A)< 1}$, the quotient $\lim_{\text{Dom}(f) \ni A_n \to A} \frac{ |f(A_n) - f(A)|}{|A_n - A|}$ exists? – MyCindy2012 Mar 10 '19 at 00:27
  • It is not clear to me what the derivative would even mean. Usually when $f:\mathbb{R}^n\to\mathbb{R}$ (Note that $M_n(\mathbb{R})$ is basically $\mathbb{R}^{n^2}$) has a derivative at a point $x$, this derivative is a linear map $Df(x):\mathbb{R}^n\to\mathbb{R}$ that gives the derivative in any direction. I believe there can even be points $A\in\mbox{Dom}(f)$ with no convex neighbourhood in $\mbox{Dom}(f)$, so I don't see how $Df(x)$ could be a linear map for such points. – SmileyCraft Mar 10 '19 at 01:04
  • May I ask how we know that there exists $B$ to guarantee that the representation of $v$ in terms of $\beta$ has only nonzero entries? Thanks. – MyCindy2012 Mar 11 '19 at 21:54
  • I was hoping this was intuitively clear, as the rigorous details are very nasty. I will give you my intuitive explanation, and if you are not satisfied with it, I will see if I can give you a (probably very nasty) formal proof.

    Heuristically speaking, if you take a matrix $B$ with complex eigenbasis $\beta$ at random (in any open set), there is a zero probability for any coordinate of $v$ w.r.t. $\beta$ to be zero. The change in spectral radius is somewhat proportional to the distance from $A$, so you can rescale $B$ in order to force $\rho(B)\geq1$ without $|A-B|$ becoming too large.

     – SmileyCraft Mar 11 '19 at 23:57
  • Thanks so much for your intuition. I do want to see a formal proof and will appreciate even if you give an outline. – MyCindy2012 Mar 12 '19 at 03:00
  • Could you point me a reference to the statement in your comment that there is a zero probability for any coordinate of $v$ w.r.t. $\beta$ to be zero? Thanks. – MyCindy2012 Mar 13 '19 at 17:24
  • I've been thinking about it some more, and I'll be honest. I don't know how to finish the technical details. I've been trying to look for references to use, but no luck yet. I believe both following claims to be true. 1. The set of matrices in $M_n(\mathbb{R})$ such that all coordinates of $v$ with respect to the eigenbasis are non-zero is dense. 2. The complement of that set has zero Lebesgue measure in $\mathbb{R}^{n^2}$. Note 2 implies 1 and 1 is enough to find $B$. Maybe someone else can find a proof or references for these facts. I also believe this becomes easier with complex matrices. – SmileyCraft Mar 15 '19 at 19:02
  • Thanks. This is very helpful for me. – MyCindy2012 Mar 15 '19 at 19:09