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[\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)\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?

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 $N$?

I'm trying to find the geodesic that connects the identity with some circulant, symmetric matrix $U\in\mathrm{GL}(2N,\mathbb{R})$, 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[\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)\right]\,. \end{align} 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?

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[\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)\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?