I'm trying to find the geodesic that connects the identity with some circulant, symmetric matrix $U\in\mathrm{GL}(N,\mathbb{R})$, meaning we have
\begin{align}
 U=\left(\begin{array}{ccc}
 u\_1 & u\_2 & \cdots\\
 u\_2 & u\_1 & \cdots\\
 \vdots & \vdots & \ddots
 \end{array}\right)\quad\text{with}\quad b\_{N-i}=b\_{1+i}\,.
\end{align}
Given two matrices $A$ and $B$ in the Lie algebra $\mathrm{gl}(2N,\mathbb{R})$, their inner product is defined by
\begin{align}
g(A,B)=\mathrm{tr}(\left(\begin{array}{ccccc}
 a\_{11} & \alpha\,a\_{12} & \cdots & \alpha^2\,a\_{1(N-1)}& \alpha\,a\_{1N}\\
 \alpha\,a\_{21} & a\_{22} & \cdots & \alpha^3\,a\_{2(N-1)}& \alpha^2\,a\_{1N}\\
 \vdots & \vdots & \ddots & \vdots & \vdots
 \end{array}\right)\left(\begin{array}{ccc}
 b\_{11} & \alpha\,b\_{12} & \cdots & \alpha^2\,b\_{1(N-1)}& \alpha\,b\_{1N}\\
 \alpha\,b\_{21} & b\_{22} & \cdots & \alpha^3\,b\_{2(N-1)}& \alpha^2\,b\_{1N}\\
 \vdots & \vdots & \ddots & \vdots & \vdots
 \end{array}\right))
\end{align}
where $\alpha$ is a positive real number. For other tangent vectors, we require that the metric is right-invariant.


The metric has the same invariance as circulant & symmetric matrices. I want to argue that the geodesic itself $U(t)$ is itself circulant & symmetric for every $t$. Is there a simple way to prove this?


**After Robert's very helpful answer, I kept thinking about the following:**
My $U$ is actually defined by the condition $UU^\intercal=G$ where $G$ is a positive definite, circulant, symmetric matrix. Intuitively, I just thought that $U$ should be itself circulant and symmetry and therefore $U=\sqrt{G}$. However, in general there is full class of matrices $U$ satifying $UU^\intercal=G$, namely $U\to U\cdot O$ where $O\in\mathrm{SO}(N)$ with $UO(UO)^\intercal=UOO^\intercal U^\intercal=UU^\intercal=G$. In this case, there is a geodesic from $1\!\!1$ to $U$ for every $U$ in this set. I still want to argue that the $U$ from above is the one with the shortest distance and I could verify this explicitly for the case $N=2$. Is there a simple argument that generalizes this to larger $N$?



Geodesic in space of circulant matrices
The answer is as follows: When $U$ is a positive-definite, symmetric, circulant matrix in $\mathrm{GL}(n,\mathbb{R})$, then there is a symmetric circulant matrix $u$ such that $U = e^u$ and the curve $\gamma(t) = e^{tu}$ (which is a positive-definite, symmetric, circulant matrix for all $t$) is a geodesic in the metric $g\_\alpha$ on $\mathrm{GL}(n,\mathbb{R})$ described by the OP for each $\alpha>0$.


Here is a sketch of a proof.


First, fix $\alpha>0$ and define a vector space isomorphism $\phi:{\frak{gl}}(N,\mathbb{R})\to {\frak{gl}}(N,\mathbb{R})$ by $\phi(A\_{ij}) = A'\_{ij}$ where $A'\_{ij} = \alpha^{d(i,j)}A\_{ij}$ and where $d(i,j)=d(j,i)\ge0$ is the mininum distance from $i-j$ to an integer multiple of $N$. Let $g\_\alpha$ be the right-invariant (pseudo-)Riemannian metric on $\mathrm{GL}(N,\mathbb{R})$ for which the inner product at the identity $I\_N\in\mathrm{GL}(N,\mathbb{R})$ is 
$$
g\_\alpha(A,B) = \mathrm{tr}\bigl(\phi(A)\phi(B)\bigr) 
= \mathrm{tr}\bigl(\phi^2(A)B\bigr)= \mathrm{tr}\bigl(A\phi^2(B)\bigr).
$$
(Note that $g\_\alpha$ is both left- and right- invariant if and only if $\alpha = 1$.)


Let $C\_N$ denote the (vector) space of symmetric, circulant $N$-by-$N$ matrices. Then $C\_N$ is closed under multiplication and forms a commutative sub-ring of the matrix ring $M\_N(\mathbb{R})$ and an abelian subagebra of the Lie algebra ${\frak{gl}}(N,\mathbb{R})$. Let $C\_N^+\subset C\_N$ denote the set of positive definite elements of $C\_N$, so that $C\_N^+$ is the connected component of $C\_N\cap \mathrm{GL}(N,\mathbb{R})$ that contains $I\_n$. The usual matrix exponential mapping $\mathrm{exp}(u) = e^u$ induces a bijective diffeomorphism $\mathrm{exp}:C\_N\to C\_N^+$. Also, note that $\phi(C\_N) = C\_N$.


Meanwhile, calculation shows that, $\gamma:\mathbb{R}\to \mathrm{GL}(N,\mathbb{R})$ is a geodesic in the metric $g\_\alpha$ if and only if $v(t) = \gamma'(t)\gamma(t)^{-1}$ satisfies the *Euler equation* for $g\_\alpha$, namely,
$$
v'(t) = - \phi^{-2}\bigl(\bigl[v(t),\phi^2(v(t))\bigr]\bigr).
$$


Now, consider the curve $\gamma(t) = e^{tu}$ for $u\in C\_n$. It satisfies 
$v(t) = \gamma'(t)\gamma(t)^{-1} = u$, and, since both $u$ and $\phi^2(u)$ belong to $C\_N$, it follows that 
$$
0 = v'(t) = - \phi^{-2}\bigl(\bigl[u,\phi^2(u)\bigr]\bigr) = -\phi^{-2}(0) = 0.
$$
Thus, $\gamma:\mathbb{R}\to \mathrm{GL}(N,\mathbb{R})$ is a geodesic for all $u\in C\_N$, as claimed.


