The way I have always been told to check if something is a manifold (I haven't had a whole lot of experience with them), is to check if the derivative of the function representing the loci of the graph satisfies the implicit function theorem. For example, the circle is a manifold because it can be expressed as $F(x,y)=x^2+y^2-1=0$ and this has a derivative of $D_F=[2x\ 2y]$ which satisfies the implicit function theorem since it is onto.
However, how could one show something more abstract is a manifold? Specifically, how can you show (even just hints are fine) that the set of all 2x3 matrices with rank one (subset of $\text{Mat}(2, 3)$) form a manifold in $\mathbb{R}^4$?
Also, the notion of manifold for me I think is properly called an embedded submanifold or something of the sort
I would like to elaborate on the nice suggestion provided by @user8268. The general fact is that the rank $k$ matrices in $Mat(m,n)$ are precisely those that are equivalent via row and column operations to the $m\times n$ matrix with an upper-left block of $I_k$ and $0$ elsewhere. The row operations are implemented by the $GL(m)$-action, while the column operations are implemented by the $GL(n)$-action. Hence, the matrices of rank $k$ are precisely those in the $GL(m)\times GL(n)$-orbit of the matrix I mentioned earlier.
Note that $GL(m)\times GL(n)$ is a real algebraic group acting algebraically on $Mat(m,n)$. Therefore, an orbit of this action is automatically a smooth locally closed subvariety of $Mat(m,n)$. In particular, any orbit is a locally closed submanifold. However, it need not be a closed submanifold.
Your example not a closed submanifold. One way to see this is to look at its defining equations. You have the equations corresponding to the vanishing of the determinants of all submatrices of size $2$. To obtain those matrices that are of rank exactly $1$, you must remove the zero-matrix from the vanishing locus of the above equations. Yet, the zero-matrix lies in the closure of the rank-$1$ matrices.