Let $F$ be a bounded subset of ${\bf M}\_n({\mathbb C})$. G.-C. Rota & G. Strang defined the joint spectral radius of $F$ as follows. For $k\ge1$, denote $F\_k$ the set of all products of $k$ elements of $F$. Set $\|F\_k\|$ the supremum of some matrix norm over $F\_k$. The sequence $\|F\_k\|^{\frac1k}$ converges to its lower bound, denoted $\rho(F)$. This joint spectral radius does not depend upon the choice of the norm.



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> I am interested in the case where $F$ consists in two elements $M$ and $M^\*$. I suspect that $\rho(F)$ equals $\rho(M^\*M)^{\frac12}$, that is $\|M\|\_2$ where $\|\cdot\|\_2$ is the standard operator norm over the Hermitian space ${\mathbb C}^n$.
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I am especially interested in the case where $F=\{L,U\}$ with 
$$L=\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix},\qquad U=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.$$
This example occurs in the study of the set $S$ of $2\times 2$ matrices with entries in $\mathbb N$ and of unit determinant. It is known that $S$ is the monoid generated by $F$, and that there are no relations between $L$ and $U$. In other words, every element $M\in S$ is a unique word in the alphabet $F$. If the claim above is correct, then every element of length $k$ satisfies $\|M\|\_2\le\phi^k$, where $\phi$ is the golden ratio.



Joint spectral radius of $\{M,M^T\}$
Your suspicion is correct. The lower bound is obtained by simply alternating $M$ and its transpose. The upper bound follows from sub-multiplicativity of the norm and the fact that 
the operator norm $\|M\|\_2$ equals the top singular value of $M$, namely $\rho(M^\*M)^{\frac12}$.


