Let $\mathbb{T}(n;\mathbb{R})\subset\mathbb{M}(n;\mathbb{R})$ denote the set of all matrices whose trace is zero. How do I
find the quotient space $\mathbb{M}(n;\mathbb{R})/\mathbb{T}(n;\mathbb{R})$, upto isomoprhism?
As much as I have understood $\mathbb{M}(n;\mathbb{R})\cong\mathbb{R}^{n^2}$, and as for any matrix in $\mathbb{T}(n;\mathbb{R})$, the one of the diagonal element will be determined by the sum of the rest of the diagonal elements, $\mathbb{T}(n;\mathbb{R})\cong\mathbb{R}^{n^2-1}$. Hence the quotient space $\mathbb{M}(n;\mathbb{R})/\mathbb{T}(n;\mathbb{R})\cong\mathbb{R}^{n^2}/\mathbb{R}^{n^2-1}\cong\mathbb{R}$.
Have I done it right?
I'm not quite sure what you mean by “find”, but yes, the quotient space is one-dimensional.
The trace itself is a linear map from your $\mathbb{M}(n;\mathbb{R})$ onto $\mathbb{R}$. Its kernel or nullspace is, by definition, $\mathbb{T}(n;\mathbb{R})$. So by the first isomorphism theorem, $$ \mathbb{M}(n;\mathbb{R})/\mathbb{T}(n;\mathbb{R}) \cong \mathbb{R} $$