What is the limit of $\mathrm{Tr}(G^kM{G^*}^k)^{1/2k}$ when $k$ goes to infinity? If $G\in \mathscr M_n(\mathbf C)$ then it's well known that $\lim_{k\to \infty}\|G^k\|^{1/k}=\rho(G)$ where $\rho(G)$ is the spectral radius of $G$, the value of the limit does not depend on the choosen norm. Consequently if we take the Schur norm we get
$$
\lim_{k\to \infty} \mathrm{Tr}(G^k{G^*}^k)^{1/2k}=\rho(G).
$$ 
Now if $M$ is an hermitian positive definite matrix then $\|A\|_M=\sqrt{\mathrm{Tr}(AMA^*)}$ is still a norm on $\mathscr M_n(\mathbf C)$ and so we also get
$$
\lim_{k\to \infty} \mathrm{Tr}(G^kM{G^*}^k)^{1/2k}=\rho(G).
$$ 
I would like to know what happen when $M$ is only hermitian positive semi-definite.  More precisely my question is the following.

Question :
  Let $G\in \mathscr M_n(\mathbf C)$ be an invertible
  matrix and let $M$ be an hermitian positive semi-definite matrix with $\mathrm{Tr}(M)=1$. Is it true that the limit
  $$\lim_{k\to \infty}\mathrm{Tr}(G^kM{G^*}^k)^{1/2k}$$
  exists and is always equal to the modulus of an eigenvalue of $G$ ?

I know that the $\lim \inf$ is larger than the smallest singular value and that the $\lim\sup$ is smaller than the largest singular value. 
 A: EDIT. I think that your new conjecture is true. 
$\textbf{Proposition 1}$. Let $G\in GL_n(\mathbb{C})$ and $M$ be a Hermitian $\geq 0$ $n\times n$ matrix with $rank(M)=p$. Let $spectrum(G)=(\lambda_i)$ where $|\lambda_1|\geq\cdots\geq |\lambda_n|$.
Then $\lim_{k\to \infty} \mathrm{tr}(G^kM{G^*}^k)^{1/2k}=|\lambda_i|$ for some $i$ s.t. $1\leq i\leq n-p+1$.
$\textbf{Proof}$. I don't write some details. $(e_i)_i$ denotes the canonical basis of $\mathbb{C}^n$.
An easy lemma
$\textbf{Lemma}$. i) If $M\leq N$, then $\mathrm{tr}(G^kM{G^*}^k)^{1/2k}\leq \mathrm{tr}(G^kN{G^*}^k)^{1/2k}$.
ii) If $M$ is changed with $\alpha M$ ($\alpha>0$), then the studied limit is the same.
We may assume that $M=diag(m_1,\cdots,m_p,0_{n-p})$; according to the lemma, we may assume that the $m_i$ are $1$ ($diag(1/2,1/2,0,0)\leq M=diag(1/2,2,0,0)\leq diag(2,2,0,0)$
and we multiply by $2$). In particular, $M^2=M$.
$\mathrm{tr}(G^kM{G^*}^k)=\mathrm{tr}({G^*}^kG^kM^2)=tr(M{G^*}^kG^kM)$ and 
$E_k(u)=u^*M{G^*}^kG^kMu=(Mu)^*({G^*}^kG^k)(Mu)$. Note that $Mu$ goes through $span(e_1,\cdots,e_p)$. 
Let $(v_i)_i$ be an orthonormal basis of eigenvectors of ${G^*}^kG^k$ associated to the singular values ${\sigma_1}^2(G^k)\geq \cdots\geq{\sigma_n}^2(G^k)$. 
$Z=span(e_1,\cdots,e_p)\cap span(v_1,\cdots,v_{n-p+1})$ is a vector-space of dimension $\geq 1$. 
If $Mu\in Z\setminus\{0\}$, then $Mu=\sum_{i\leq n-p+1}a_iv_i$ and 
$${\sigma_{n-p+1}}^2(G^k)\leq \dfrac{E_k(u)}{||Mu||^2}=\dfrac{\sum_{i\leq n-p+1}\sigma_i^2(G^k)|a_i|^2}{\sum_{i\leq n-p+1}|a_i|^2}\leq {\sigma_1}^2(G^k).$$
Let $f(u)=\min\{i;a_i\not= 0\}$ and $j=min_u f(u)$.
Then $j$ is constant for $k$ great enough (to check) and $\dfrac{E_k(u)}{||Mu||^2}$ behaves essentially like ${\sigma_j}^2(G^k)$ (up to a factor) -to check-.
Note that, if the $(b_i)$'s are $>0$, then $\lim_{k\to \infty} (b_1^{2k}+\cdots+b_n^{2k})^{1/2k}=\sup(b_1,\cdots,b_n)$.
Finally, roughly speaking, we seek
$\lim_{k\to \infty} ({\sigma_j}^2(G^k))^{1/2k}=\lim_{k\to \infty} (\sigma_j(G^k))^{1/k}$.
According to a result by Yamamoto, the above limit is $|\lambda_j|$.  $\square$
$\textbf{Proposition 2}.$ For a generic matrix $G$,
$\lim_{k\rightarrow +\infty}(tr(G^kMG^{*k}))^{1/2k}=\rho(G)$.
$\textbf{Proof}$. Consider a generic $G$ (for example, randomly choose it). 
Up to an orthonormal change of basis, we may assume that $M=diag(m_1,\cdots,m_p,0_{n-p})$ where $m_i>0$ and $p\geq 1$.
Let $u=(u_i)$ be a unitary vector s.t. $G^*u=\overline{\lambda} u$ where $|\lambda|=\rho(G)$. Generically, $u$ is unique (up to a factor) and $G$ is diagonalizable.
Then $E(u)=u^*G^kMG^{*k}u=u^*\lambda^k(\overline{\lambda^k} Mu)=\rho(G)^{2k}u^*Mu$ and
$E(u)=\rho(G)^{2k}\sum_i m_i|u_i|^2$.
Generically, at least one of the $u_i$ is non-zero (in fact all the $u_i$) and  
$E(u)\sim a\rho(G)^{2k}$ when $k\rightarrow +\infty$ where $a=\sum_i m_i|u_i|^2$.
If $v$ is another unitary vector, then 
$E(v)=O(||G^{*k}||||G^k||)=O(\rho(G)^{2k})$.
Then, for $k$ great enough, $tr(G^kMG^{*k})\geq \dfrac{a}{2}\rho(G)^{2k}$ and  $tr(G^kMG^{*k})=O(\rho(G)^{2k})$ and we are done. $\square$
Remark. if you choose (for example) $G$ and $M$ as diagonal matrices, then, of course, we may find another limit than the one above.  
A: Here is a slightly different answer based on several ideas of loup blanc. I think that the advantage of this answer is that we chose to decompose the vectors over a Jordan basis of $G$ which does not depend on $k$. 
Let $(e_1,\ldots,e_n)$ be the canonical basis of $\mathbf C^n$ and $(\cdot, \cdot)$ be the standard Hermitian product. We then have $\mathrm{Tr}(A)=\sum_{i=1}^n(Ae_i,e_i)$ for every hermitian matrix $A$. If $P$ is a unitary matrix then $$\mathrm{Tr}(A)=\mathrm{Tr}(P^*AP)=\sum_{i=1}^n(P^*APe_i,e_i)=\sum_{i=1}^n(APe_i,Pe_i).$$
So for any orthonormal basis $(f_1,\ldots, f_n)$ and any hermitian matrix $A$ we have
$$\mathrm{Tr}(A)=\sum_{i=1}^n(Af_i,f_i).$$ 
Now since the matrix $M$ is hermitian positive semi-definite there exists a matrix $N$ which is hermitian positive semi-definite such that $NN=M$. Consequently
$$\mathrm{Tr}(G^kM{G^k}^*)=\mathrm{Tr}(G^kNN{G^k}^*)=\mathrm{Tr}((G^kN)^*G^kN)=\sum_{i=1}^n(G^kNf_i,G^kNf_i)$$
where $(f_1,\ldots,f_n)$ is any orthonormal basis. If we choose $(f_i)_i$ to be an orthonormal basis of eigenvectors of $N$ then we get
$$
\mathrm{Tr}(G^kM{G^k}^*)=\sum_{i=1}^n\mu_i^2(G^kf_i,G^kf_i) = \sum_{i=1}^n\mu_i^2\|G^kf_i\|_2^2
$$ 
where $Nf_i = \mu f_i$ and $\|\cdot\|_2$ is the euclidean norm. We finally decomposes the $f_i$ over a Jordan basis for $G$ : $f_i=\sum \alpha_j^i v_j$ and compute the $G^kf_i$. If all the $v_j$ are eigenvectors then the computation is easy. If it's not the case then fix some integer $J$ and assume that $Gv_1 = \lambda_1 v_1 $ and $Gv_j = \lambda_1 v_j + v_{j-1}$ for every $j=2,\ldots,J$. We then have
$$G^k v_J = \lambda_1^k v_J + \binom{k}{1}\lambda_1^{k-1}v_{J-1} + \ldots + \binom{k}{J-1}\lambda_1^{k-J+1}v_1=\lambda^{k-J+1}\binom{k}{J-1}(v_1+ o(1))$$
and for the same reason if $\beta_J \in \mathbf C \backslash \{0\} $ then
$$
G^k(\beta_1 v_1 + \ldots + \beta_J v_J) = \beta_J\lambda_1^{k-J+1}\binom{k}{J-1}(v_1+ o(1)).
$$
Consequently there exists some eigenvector $v_{i_1}$ of $G$, some non zero complexe number $\beta_{i_1}$ and some integer $N_{i_1}$ such that 
$$G^k f_1 = \beta_{i_1}\lambda_{i_1}^{k-N_{i_1}+1}\binom{k}{N_{i_1}}(v_{i_1}+o(1)).$$
This is enough to conclude using $\lim (|\lambda_1|^k+\ldots+|\lambda_n|^k)^{1/k}=\sup_{i=1,\ldots,n} |\lambda_i|$ and well known results on the comparative growth of functions :
$$
\lim_{k\to\infty} \mathrm{Tr}(G^kM{G^k}^*)^{1/2k}=|\lambda_i|
$$
where $\lambda_i$ is some eigenvalue of $G$.
