I had already asked this [question in MSE](https://math.stackexchange.com/questions/2408030/is-a-linear-vector-field-a-geodesible-vector-field) then I ask here at MO

Assume that $A\in M_n(\mathbb{R})$ 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.

2) $A^2$ 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 $6.8$ 


https://mathoverflow.net/questions/273635/finding-a-1-form-adapted-to-a-smooth-flow/273648#273648
 

Please see also this related post:

https://mathoverflow.net/questions/282694/is-every-real-matrix-conjugate-to-a-semi-antisymmetric-matrix