I am trying to prove that the given map below is a diffeomorphism and it is pretty clear to me that it is a bijection but I do not know how to show that the inverse of the given map is smooth? in fact I do not know how to find its inverse (on its image), could someone help me in this please?
Prove that $$\mathbb C \to M_2(\mathbb R),$$ $$x + iy \mapsto \begin{pmatrix} x & -y\\ y & x \end{pmatrix}$$ defines an imbedding.
EDIT:
Is the inverse can be found by finding the inverse of the given matrix?
EDIT: 2
Here is the question I am trying to solve:
Prove that $$\mathbb C \to M_2(\mathbb R),$$ $$x + iy \mapsto \begin{pmatrix} x & -y\\ y & x \end{pmatrix}$$ defines an imbedding. More generally it defines an imbedding $$M_n(\mathbb C) \to M_n(M_2(\mathbb R)) \cong M_{2n}(\mathbb R).$$ Show also that this imbedding sends "conjugate transpose" to "transpose" and "multiplication" to "multiplication".
I am trying to prove that the given map is a diffeomorphism which means that it is a bijection and its inverse is smooth.
Could you please tell me if the question has a typo or no?
Thanks!!
EDIT:
Define the function $f$ as $$x + iy \mapsto \begin{pmatrix} x & -y\\ y & x \end{pmatrix}$$
and define the set $$\mathcal{M}:=\left\{\begin{bmatrix}x & -y\\y & x\end{bmatrix}:x,y\in\mathbb{R}\right\}.$$
It is immediate to see that $f(\mathbb{C})=\mathcal{M}\subset M_2(\mathbb{R})$.
It is clear then that the map $f$ you are considering is a bijection from $\mathbb C$ to $\mathcal{M}$ and that this function is differentiable.
The function $f$ has then an inverse $g:\mathcal{M}\mapsto\mathbb{C}$ which may take many forms but one simple one is
$$g(M)=e_1^TMe_1+ie_2^TMe_1$$
where $\{e_1,e_2\}$ is the natural basis for $\mathbb{R}^2$. This function is easily seen to be differentiable as it is affine in $M$.
We also have that $f(z^*)=f(z)^T$, $f(z_1+z_2)=f(z_1)+f(z_2)$, and $f(z_1z_2)=f(z_1)f(z_2)$ which can be verified by direct evaluation.
Now define the map $h:M_n(\mathbb C)\mapsto M_n(M_2(\mathbb R))$ as
$$h(M) = \sum_{i,j=1}^ne_ie_j^T\otimes f(m_{ij})$$
where $\otimes$ denotes the Kronecker product and $\{e_i\}$ is here the standard basis for $\mathbb{R}^n$. This is again an embedding (for the same reasons as for $f$) and we also have that
$$h(M^*)=\sum_{i,j=1}^ne_je_i^T\otimes f(m_{ij})^T= \left(\sum_{i,j=1}^ne_ie_j^T\otimes f(m_{ij})\right)^T=h(M)^T$$
where we have used the fact that the transpose of the sum of the sum of the transpose, and that for two matrices $A$ and $B$, the following equality $(A\otimes B)^T=A^T\otimes B^T$ holds. This shows that the embedding sends "transpose conjugation" to "transpose".
Now, let $C=AB$ for $A,B\in M_n(\mathbb C)$. We have that
$$h(C)=\sum_{i,j=1}^ne_ie_j^T\otimes f(c_{ij}).$$
But $$f(c_{ij})=f\left(\sum_{k=1}^na_{ik}b_{kj}\right)=\sum_{k=1}^nf\left(a_{ik}b_{kj}\right)=\sum_{k=1}^nf(a_{ik})f(b_{kj}),$$ where we have used the additive and multiplicative properties of $f$ obtained above. Therefore,
$$h(C)=\sum_{i,j=1}^ne_ie_j^T\otimes \left(\sum_{k=1}^nf(a_{ik})f(b_{kj})\right)=\sum_{i,j,k=1}^ne_ie_j^T\otimes (f(a_{ik})f(b_{kj})).$$
If we evaluate now $h(A)h(B)$, we get that
\begin{equation} \begin{array}{rcl} h(A)h(B)&=&\left(\sum_{i,j=1}^ne_ie_j^T\otimes f(a_{ij})\right)\left(\sum_{i,j=1}^ne_ie_j^T\otimes f(b_{ij})\right)\\ % &=&\sum_{i,j,k,\ell=1}^n\left(e_ie_k^T\otimes f(a_{ik})\right)\left(e_\ell e_j^T\otimes f(b_{\ell j})\right)\\ % &=&\sum_{i,j,k,\ell=1}^n\left(e_ie_k^Te_\ell e_j^T\right)\otimes\left(f(a_{ik})f(b_{\ell j})\right)\\ % &=&\sum_{i,j,k=1}^ne_ie_j^T\otimes\left(f(a_{ik})f(b_{kj})\right)\\ &=&h(AB) \end{array} \end{equation}
This, finally, proves that this embedding sends "multiplication" to "multiplication".