Seeking Left $SL(n, \mathbb{R})$-Invariant Canonicalization for Full-Rank Square Matrices

Question

Question: Let $P\in \mathbb{R}^{n\times n}$ be a full-rank square matrix. I am interested in determining whether it is possible to define a canonicalization mapping $c:\mathbb{R}^{n\times n} \rightarrow \mathbb{R}^{n\times n}$ such that

1. Mapping $c(P)$ remains invariant under any left multiplication by matrices in $SL(n,\mathbb{R})$. In other words, for any invertible matrix $A\in \mathbb{R}^{n\times n}$, $\text{det}(A) = 1$, we have $c(AP) = c(P).$
2. There exists $A \in SL(n,\mathbb{R})$ such that $c(P) = AP$.

I am only aware that if $GL(n,\mathbb{R})/SL(n,\mathbb{R})$ is acting on $P$ by conjugate multiplication, the canonical form of $P$ can be written as Jordan normal form or Weyr form in specific block order. But left invariant by $SL(n,\mathbb{R})$ is challenging as eigenvalues or singular values are not left $SL(n, \mathbb{R})$ invariant.

Any insights, references, or suggestions for further reading would be greatly appreciated.

Answer 1

$\mathrm{SL}(n,\mathbb{R})$ acts transitively on subspaces of a given dimension, so for any $P$ there is an $A\in\mathrm{SL}(n,\mathbb{R})$ which sends $\mathrm{img}(P)$ to $\ker(P)^\perp$ (same dimension by rank-nullity). Then $AP$ acts as $0$ on $\ker(P)$ and as an invertible operator $P'$ on $\ker(P)^\perp$. Note $\det(P')$ is well-defined (independent of $A$). Pick a $1$D subspace $L\le\ker(P)^\perp$. Let $c(P)$ act as $\det(P')$ on $L$, by $I$ on $L^\perp$ within $\ker(P)^\perp$, and by $0$ on $\ker(P)$.

So $c(-)$ is a section of the projection from $\mathbb{R}^{n\times n}$ to its orbit space under left multiplication by $\mathrm{SL}(n,\mathbb{R})$. This requires choosing a $1$D subspace of every possible subspace, which I don't see a way around.