Question about Isomorphism I was reading this example on Wikipedia.
It says the following:
Another example is the quotient of $\mathbb{R}^n$ by the subspace spanned by the first $m$ standard basis vectors. The space $\mathbb{R}^n$ consists of all $n$-tuples of real numbers $(x_1, ..., x_n)$. The subspace, identified with $\mathbb{R}^m$, consists of all $n$-tuples such that the last $n-m$ entries are zero: $(x_1, ..., x_m, 0, 0, ..., 0)$. Two vectors of $\mathbb{R}^n$ are in the same congruence class modulo the subspace if and only if they are identical in the last $n-m$ coordinates. The quotient space $\mathbb{R}^n/\mathbb{R}^m$ is isomorphic to $\mathbb{R}^{n-m}$ in an obvious manner.
I don't understand the last line. Why is $\mathbb{R}^n/\mathbb{R}^m$ is isomorphic to $\mathbb{R}^{n-m}$? Can anyone please help? Maybe I am not clear on what is an isomorphism. Can anyone explain?
Thanks!

You can write the quotient space as $\mathbb{R}^n/\mathbb{R}^m$={v + $\mathbb{R}^m$} with v $\in$ $\mathbb{R}^n$.  When we consider the projection p from $\mathbb{R}^n$ to $\mathbb{R}^n/\mathbb{R}^m$, we can see that its kernel is: ker(p)={v $\in$ $\mathbb{R}^m$}, since p(v)=0 for all v $\in$  $\mathbb{R}^m$. Therefore, it the dimension of $\mathbb{R}^n/\mathbb{R}^m$ has to be n-m, since dim(ker(p))+dim(im(p))=n and dim(ker(p))=m.
