Given a real matrix $W$, find a complex matrix $V$ such that $df_A(V) = W$ where $df_A: V \mapsto A\bar{V}^T + V\bar{A}^T, A\bar{A}^T = 1_{n\times n}$ I am trying to complete the proof that the unitary group is a smooth manifold using regular values. To complete this proof I need to show that for any real-valued matrix $W$, there exists a complex-valued matrix $V$ such that
$df_A(V) = W$ where $df_A: V \mapsto A\bar{V}^T + V\bar{A}^T$.
where $A$ is a complex matrix such that $A\bar{A}^T = 1_{n\times n}$ (here $\bar{A}$ denotes the complex conjugate of $A$, and $A^T$ denotes the transpose of $A$).
I originally thought to use $V = \frac{1}{2}WA$, but this gives $df_A(V) = \frac{1}{2}W + \frac{1}{2}W^T$ which is unfortunately not $W$. Any ideas on a $V$ that would satisfy the above equation?
Any help is much appreciated

Let me denote $\overline{A}^T$ by $A^{*}$. I guess you are trying to describe $U(n)$ as the inverse image $f^{-1}(I_n)$ for $f \colon M_n(\mathbb{C}) \rightarrow M_n(\mathbb{C})$ given by $f(A) = A \cdot A^{*}$. We indeed have
$$ df|_A(V) = AV^{*} + VA^{*} $$
but note that $(AV^{*} + VA^{*})^{*} = VA^{*} + AV^{*} = AV^{*} + VA^{*}$ so $df|_A(V)$ is always a Hermitian matrix and so no value can be regular for $f$. The reason is that the codomain of $f$ is too big - no matter what $A$ is, the matrix $A A^{*}$ will always be a Hermitian matrix so you can consider as well the same map $f$ but with the codomain restricted to the real vector space of $n \times n$ Hermitian matrices. If you do this, then $I_n$ will be a regular value for $f$. For that, you will need to show that for each Hermitian matrix $W$ and a unitary matrix $A$, you can find a complex matrix $V$ such that $AV^{*} + VA^{*} = W$. Since
$$ df|_A\left( \frac{WA}{2} \right) = \frac{A(WA)^{*} + (WA)A^{*}}{2} = \frac{W^{*} + W}{2} = W $$
we see that indeed $I_n$ is a regular value for $f$.
