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Math Help - homeomorphism between R^(n^2) and M_n(R)

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    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by oblixps View Post
    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?
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    ah i see, thanks! isometries are continuous and since f is a surjective isometry the inverse of f must be continuous as well correct?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by oblixps View Post
    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.
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