# Thread: homeomorphism between R^(n^2) and M_n(R)

1. ## homeomorphism between R^(n^2) and M_n(R)

how do you show that M_n(R) and R^(n^2) are homeomorphic? i defined a function f:M_n(R) -> R^(n^2) such that if you input any real matrix, you get out a vector that is set up as (a11, a12, ... a21, a22, ..., a_nn) so basically i go along the first row then the second row then 3rd row and so on of the matrix and put the elements in the order i get to them while doing this process. so i can see that this function is a bijection and now i need to show that both the function f and f^-1 are continuous and i am a bit lost on how to do that.

2. Originally Posted by oblixps
how do you show that M_n(R) and R^(n^2) are homeomorphic? i defined a function f:M_n(R) -> R^(n^2) such that if you input any real matrix, you get out a vector that is set up as (a11, a12, ... a21, a22, ..., a_nn) so basically i go along the first row then the second row then 3rd row and so on of the matrix and put the elements in the order i get to them while doing this process. so i can see that this function is a bijection and now i need to show that both the function f and f^-1 are continuous and i am a bit lost on how to do that.

I don't understand the problem here? I assume that you're endowing $\text{Mat}_n\left(\mathbb{R}\right)$ with the topology induced by the usual matrix norm $\displaystyle |||A|||=\sqrt{\sum_{i=1}^{n}\sum_{j=1}|a_{i,j}|^2} ,\quad A=[a_{i,j}]$. If so, isn't

$\displaystyle f:\text{Mat}_n\left(\mathbb{R}\right)\to\mathbb{R} ^{n^2}:\begin{pmatrix}a_{1,1} & \cdots & a_{1,n}\\ \vdots & \ddots & \vdots\\ a_{n,1} & \cdots & a_{n,n}\end{pmatrix}\mapsto \left(a_{1,1},\cdots,a_{n,1},\cdots,a_{1,n} & \cdots a_{n,n}\right)$

a surjective isometry, and thus trivially a homeomorphism?

3. ah i see, thanks! isometries are continuous and since f is a surjective isometry the inverse of f must be continuous as well correct?

4. Originally Posted by oblixps
ah i see, thanks! isometries are continuous and since f is a surjective isometry the inverse of f must be continuous as well correct?
Right! In general every isometry is uniform continuous since it's Lipschitz. More specifically if $f:\left(X,d_X\right)\to\left(Y,d_Y\right)$ is an isometry then it's continuous since

given any $x_0$ and $\varepsilon>0$ we see that $f\left(B_{\varepsilon}(x)\right)\subseteq B_{\varepsilon}\left(f(x)\right)$ (in fact they're equal) since $d_X(x_0,x)<\varepsilon\implies d_Y(f(x),f(x_0))=d_X(x,x_0)<\varepsilon$.

Isometries are also always injective since if $x\ne x'$ then $0\ne d_X(x,x')=d_Y(f(x),f(x'))$ and thus $f(x)\ne f(x')$. Thus, if $f$ is a surjective isometry it

must be bijective. But, we can also see that $f^{-1}$ is an isometry. Indeed, if $y,y'\in Y$ then by surjectivity we know that $y=f(x),y'=f(x')$ for some

$x,x'\in X$ and thus $d_Y\left(y,y'\right)=d_Y(f(x),f(x'))=d_X(x,x')=d_X \left(f^{-1}(y),f^{-1}(y')\right)$ from where the conclusion follows.