I am wondering how I would find an isomorphism from a set of matrices of dimension $n\times n$ with a specific value for the trace. For simplicity, let us say the trace must be equal to $1$, although this could be generalized to any value of $x$.
I understand that this places a restriction on the matrix, such that a matrix $A \in M_{n,n}$ would need to look like the following: \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & 1 - \left( \sum_{i=1}^{n-1} a_{ii} \right) \end{bmatrix}
As far as I can tell, I then have $n^2-n$ free elements off of the diagonal, and $n-1$ free elements on the diagonal. I believe that I could then create an isomorphism to $\mathbb{R}^{n^2-1}$.
I think my challenge here is that $M_{n,n}$ is not a vector space. I am then struggling with the idea of dimensionality. Does $M_{n,n}$ simply have dimension $n^2-1$?
If $x\neq0$, then the set of matrices $T_x=\{A\in M_n(\mathbb{R})\mid\operatorname{tr}(A)=x\}$ is not a subspace but the formula $T_x=A+T_0$ for some matrix $A$ with trace $x$ is satisfied. Hence, we can assume that $\dim T_x=\dim T_0=n^2-1$, since we know that $T_0$ is a subspace in $M_n(\mathbb{R})$.
In this case, I don't think it makes sense to talk about isomorphism $T_x$ and $T_0$. In any case it is necessary to understand what you want from this isomorphism. What structure should it preserve?