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

• Jan 20th 2011, 03:30 PM
oblixps
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.
• Jan 20th 2011, 03:35 PM
Drexel28
Quote:

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

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

a surjective isometry, and thus trivially a homeomorphism?
• Jan 21st 2011, 05:02 PM
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?
• Jan 21st 2011, 05:20 PM
Drexel28
Quote:

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 $\displaystyle f:\left(X,d_X\right)\to\left(Y,d_Y\right)$ is an isometry then it's continuous since

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

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

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

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