**Edit:** According to comment by Leo Monsaingeon I revise my question:


Is there a Riemannian metric on $M\_n(\mathbb{R})$ for which the function $trace$ is a bounded function on **every complete(whole)** geodesic? that is for every **complete** geodesic $\gamma$, the restriction $trace$ to $\gamma$ is a bounded function? What about the same question if we replace $M\_n(\mathbb{R})$ with $GL\_n(\mathbb{R})$?


**Remark** By a complete geodesic we DO NOT mean exclusively a geodesic whose parameter $t$ can be varied in whole $\mathbb{R}$, but we mean $\{\gamma(t)\mid t\in I\}$ where $I$ is the maximal interval of definition of $\gamma$. So we mean the whole trajectory of a non parametrized geodesic. 



Riemannian metrics on matrix space for which the restriction of trace function to each complete geodesic is a bounded function
It is possible to find such metrics. However, they are a bit unnatural, in that they aren't really constructed from any inherent properties of matrices. For concreteness, I'm going to restrict my attention to $M\_2$ and $Gl\_2$, but this construction can be modified to work for any larger $n$ as well.


We consider $2 \times 2$ matrices $$
\left[
\begin{array}{cc}
 a & b \\
 c & d \\
\end{array}
\right]. $$


To construct the desired metric, we start by changing coordinates on $M\_2$ as follows.


$x = a+d$, $y= \exp(a-d)$, $z=b$, $w=c$


This gives a map from $M\_2$ to a half-space in $\mathbb{R}^4$. For a Riemannian metric, we use $$ds^2 = \frac{dx^2+dy^2}{y^2} +dz^2+dw^2.$$


Notice that the $x$-coordinate is simply the trace of the original matrix, so what's left to show is that $x$ is bounded on any geodesic. Different geodesics will have different bounds, however.


To show that $x$ is bounded on any geodesic, observe that as a Riemannian manifold, $(M\_n, g)$ is simply the metric product $\mathbb{H} \times \mathbb{R}^2$. As such, the geodesics are of the form $\gamma(t) = (c(t), \ell(t))$. 
Here, $c(t)$ forms an arc of a semi-circle in $\mathbb{H} $ whose center is on the axis $y=0$. Meanwhile, $\ell(t))$ is a line in $\mathbb{R}^2$. As such, it follows the $x$ coordinate is bounded by the center and radius of $c(t)$. 


When we try to do this for $Gl\_2$, we have to deal with the fact that the space is disconnected. However, we can use a similar idea on each connected component. Namely, we change coordinates to
$x = a+d$, $y= \exp(a-d)$, $z=b$, $w=ad-bc$ and take the metric 
$ds^2 = \frac{dx^2+dy^2}{y^2} +\frac{dz^2+dw^2}{w^2}.$
This gives $Gl\_2$ the metric structure of $\mathbb{H} \times \mathbb{H}$ on each of its connected components, which also has the desired geodesic property. 


