Is there an explicit left invariant metric on the general linear group? Let $\operatorname{GL}_n^+$ be the group of real invertible matrices with positive determinant.

Can we construct an explicit formula for a metric on $\operatorname{GL}_n^+$ which is left-invariant, i.e.

$$d(A,B)=d(gA,gB) \, \,\forall A,B,g \in\operatorname{GL}_n^+$$
and which induces the standard topology on $\operatorname{GL}_n^+$.
(Without the last requirement the discrete metric will do).

Even finding a concrete example of a metric which is only "scale invariant" (i.e. $d(A,B)=d(rA,rB) \, \,\forall r \in \mathbb{R}$) will be nice; Actually, even finding a metric which is invariant under multiplication by $r=2$ seems non-trivial.
A Riemannian approach:
It's easy to prove existence of left-invariant metrics:  Just left-translate any metric on the tangent space at the identity. The problem is that this usually does not induce an explicit formula for the distance.
One can take the left translation of the standard Frobenius metric on $T_I\operatorname{GL}_n^+ \simeq M_n$, and to use its induced distance. I don't know how to compute explicitly this distance:
For any symmetric positive-definite matrix $P$ $$d(I,P)=\|\log P\|_F \tag{*},$$
where $\|\cdot \|_F$ is the Frobenius norm, and $\log P$ is the unique symmetric logarithm of the matrix $P$.
This is proved here in section 3.3.
The point is that it's easier to compute $d(A,\operatorname{SO}(n))$ than $d(A,B)$: A minimizing geodesic from a point to a submanifold must intersect that submanifold orthogonally, hence we have more constraints on its velocity; this simplifies the analysis, and makes it tractable.
It can be shown that
$$ d(A,\operatorname{SO}(n))=d(A,Q(A))=\|\log \sqrt{A^TA}\|_F,$$ where $Q(A)$ is the orthogonal polar factor of $A$. In particular for positive matrices $P$, $Q(P)=I$, so we obtain $(*)$.
As far as I know, computing the distance $d(I,X)$ for an arbitrary $X \in \operatorname{GL}_n^+$ is open.
Additional partial results:
Any such left invariant metric is determined by $f(X)=d(X,I)$, since
$$ d(A,B)=d(I,A^{-1}B)=f(A^{-1}B) \tag{1}$$
Rephrasing the requirements from a metric in terms of $f$, we see that if $d$ is given in terms of $f$ as in $(1)$, then $d$ is a metric if and only if
Positivity: $f(X)=0 \iff X=I \tag{2}$
Symmetry: $f(X)=f(X^{-1}) \tag{3}$
Triangle inequality: $f(XY) \le f(X) + f(Y) \tag{4}$
Thus, we obtained an equivalent formulation of the problem:

Find a non-negative function $f:\operatorname{GL}_n^+ \to \mathbb{R}$ satisfying requirements $(2)-(4)$.

Reduction of the problem to $\operatorname{SL}_n^+$:
Consider $f(X)=|\ln (\det X)|$.  $\, \,f$ satisfies $(3),(4)$, and $f(X)=0 \iff X \in \operatorname{SL}_n^+$.
Now, suppose we constructed a function $\tilde f:\operatorname{SL}_n^+ \to \mathbb{R}^+$ satisfying $(2)-(4)$ above.
Then, by defining
$$ \hat f(X)=f(X)+\tilde f(\frac{X}{\det(X)^{\frac{1}{n}}})$$
it is easy to see that $\hat f:\operatorname{GL}_n^+ \to \mathbb{R}$ also satisfies $(2)-(4)$ as required.

Discussion:
Is it true that in some sense the space of left-invariant metrics is "finite-dimensional"? (I refer to arbitrary metrics not just those which are induced by Riemannian metrics).
To make this notion more  precise, some care should be taken. For instance, there are ways to generate new invariant metrics from old one (e.g. by applying the map $d \to \sqrt{d}$), but for this discussion we can identify two metrics if one is a function of the other.
Edit:
I now think that this space is always infinite-dimensional. Since any left translation of a smooth norm, will induce a Finsler norm, and the space of smooth norms is not 'finite-dimensional' in any reasonable way, the space of metrics is also infinite-dimensional. (Since different Finsler norms give rise to different induced distances).

For $n=1$, $\operatorname{GL}_n^+=\mathbb{R}^{>0}$, and the fomula:
$d(x,y)=|\ln(\frac{y}{x})|$ does the job. (It is in fact induced by the Riemannian metric obtained from left translation of the standrad metric on $T_1\mathbb{R}$). The obvious problem with generalizing this to higher dimensions is that there is no global matrix logarithm on $\operatorname{GL}_n^+$, one has to choose a branch.
 A: Let $P$ be a convex polytope in $R^n$ whose interior contains $0$ and which has no nontrivial linear symmetries (i.e. if $A$ is an invertible linear map, $AP=P$ implies that $A$ is the identity). In particular, $P\ne -P$. You can easily construct such $P$ by taking a suitable simplex or a cube. Then $P$ defines a nonsymmetric norm $||\cdot||$ on 
$R^n$ for which $P$ is the unit ball, by the usual procedure: $||v||=t$, where $t\in R_+$ is such that $t^{-1}v$ is on the boundary of $P$, and setting $||0||=0$.   Using this norm we define the standard operator norm on linear endomorphisms of $R^n$:
$$
||A||= \max \{||Av||: v\in P\}.
$$
 This norm satisfies $||AB||\le ||A||\cdot ||B||$ and for every invertible matrix $A$,
$$
g(A)=\max(||A||, ||A^{-1}||)\ge 1$$ 
with equality if and only if $A=I$, the identity matrix. 
 The function $g$ is "explicit" in the sense that $||A||$ is easily computable: 
 It equals maximum of the norms $||Av_i||$, where $v_i$'s are the vertices  of $P$. 
 Now, set $f(A):= \log(g(A))$. This is your function. (All the required properties are clear.)  If it is of any use, I do not know. 
