I had already asked this question in MSE then I ask here at MO

Assume that $Ain M_n(mathbb{R})$ is a non singular matrix.

Is the flow of linear vector field $X’=AX$ a geodesible flow on $mathbb{R}^n setminus {0}$?Namely, is there a Riemannian metric on $mathbb{R}^n setminus {0}$

such that the trajectories of the linear vector field are unparametrized geodesics?

**Remark:** For $n=2$ the answer is affirmative, as we explain below:

**Fact:** A linear vector field associated to a non singular$ 2 times 2$ real matrix is a geodesible vector field on the punctured plane.

**Proof:**

Let $A$ be an invertible matrix. We denote by $X$ the linear vector field associated to $A$.

We consider two cases:

1)$A^2$ has no real eigenvalue.

- $A^2$ has real eigenvalue.

Case 1) In this case the linear vector field $Y$ associated to matrix $A^{-1}$ is transverse to $X$ on the puntured plane and satisfies $(X,Y)=0$ this obviously implies that $X$ is a geodesible vector field.

Case 2) If $A^2$ has real eigenvalue then $A$ is similar to one of the following matrices:

$$begin{pmatrix} a&0\ 0& b end{pmatrix};; begin{pmatrix} a&epsilon\ 0& a end{pmatrix} ;;begin{pmatrix} 0&b\ -b& 0 end{pmatrix} $$

For the first matrix the closed one form $psi=axdx+bydy$ satisfies $psi(X)>0$.So $X$ is a geodesible vector field. For the second matrix the $1$-form $psi=axdx+aydy$ satisfies $psi(X)>0$. For the third matrix the vector field is geodesible because we have a foliation of punctured plane by closed curve.

The reason of geodesibility of case $1$ and three matrices in case $2$ is discussed in the following post which is essentially based on page 71 of “Geometry of foliation ” by Philip Toender, Propsition $6.7$ and $6.8$

Finding a 1-form adapted to a smooth flow

Please see also this related post:

Is every real matrix conjugate to a semi antisymmetric matrix?