hermitian forms are related by linear transformations Suppose $(-,-)$ and $[-,-]$ are two positive deinifite hermitian forms on an $n$-dimensional vector space, show that there exists an invertible linear transformation $\phi$ such that $(u,v) = [\phi(u),\phi(v)]$.
Attempt: I tried to write the hermitian forms in matrix forms, that is $(v,w) = vH\overline{w}^\intercal $, and $[v,w] = vJ\overline{w}^\intercal$, with the associated matrix $H$ and $J$ of the hermitian forms, and try to relate the two matrix by a linear transformation, but I can't seem to get a concrete linear transformation to do so.
Can someone help me with this?
 A: A positive definite Hermitian matrix $H$ can always be deconstructed as
$$H = Q D Q^\dagger = A A^\dagger, ~~ \text{where } A = Q \sqrt{D} Q^\dagger$$
with $D$ being a diagonal matrix with strictly positive entries and $Q$ unitary.
Let us then decompose $H = Q D Q^\dagger$ and $J = S B S^\dagger$ and choose $\phi$ to be represented by the matrix $T = Q \left(\sqrt{D} \big/ \sqrt{B}\right) S^\dagger $ where $\sqrt{D} \big/ \sqrt{B}$ is the diagonal matrix obtained by dividing the diagonal entries of $\sqrt{D}$ by the respective ones of $\sqrt{B}$. Here it is crucial that both $H$ and $J$ are positive semi-definite, as this guarantees that the division is well-defined as none of the diagonal elements of $\sqrt{B}$ can be zero. It also further guarantees that the diagonal elements of $\sqrt{D} \big/ \sqrt{B}$ are not zero, as none of the diagonal elements of $\sqrt{D}$ can be zero. Then it follows that
\begin{align*}
[\phi(u), \phi(v)] = u T (S B S^\dagger) T^\dagger v^\dagger &= u Q \left(\sqrt{D} \big/ \sqrt{B}\right) B \left(\sqrt{D} \big/ \sqrt{B}\right) Q^\dagger v^\dagger \\ &= u Q D Q^\dagger v^\dagger = u H v^\dagger = (u, v)
\end{align*}
Note that $\left(\sqrt{D} \big/ \sqrt{B}\right) = \left(\sqrt{D} \big/ \sqrt{B}\right)^\dagger$ as it is a diagonal matrix with real entries.
It remains to check if $T$ is invertible. Indeed it is, as it is the product of three invertible matrices: $Q$ and $S^\dagger$ are unitary and hence invertible, and $\left(\sqrt{D} \big/ \sqrt{B}\right)$ as noted before is a diagonal matrix with strictly positive entries (and hence invertible). So $\phi$ is invertible, and we are done. $\square$
A: Call the vector space $V$. Let $A=(\cdot,\cdot)$ and $B=[\cdot,\cdot]$. Since $A$ is Hermitian, it is unitarily diagonalisable with respect to the inner product $B$. That is, there exist a $B$-orthonormal basis $\{u_1,u_2,\ldots,u_n\}$ and $n$ real numbers $\lambda_1,\lambda_2,\ldots,\lambda_n$ such that $A\left(\sum_ic_iu_i,\sum_ic_iu_i\right)=\sum_i\lambda_i|c_i|^2$. Since $A$ is positive definite, each $\lambda_i$ is positive. Now define $\phi(u_i)=\sqrt{\lambda_i}u_i$ for each $i$. Then
\begin{aligned}
B\left(\phi(\sum_ic_iu_i),\phi(\sum_ic_iu_i)\right)
&=B\left(\sum_i\sqrt{\lambda_i}c_iu_i,\sum_ic_i\sqrt{\lambda_i}u_i\right)\\
&=\sum_i|\sqrt{\lambda_i}c_i|^2\\
&=\sum_i\lambda_i|c_i|^2\\
&=A\left(\sum_ic_iu_i,\sum_ic_iu_i\right).
\end{aligned}
In other words, we have $B(\phi(x),\phi(x))=A(x,x)$ for every $x\in V$. It follows from the polarisation identity that $B(\phi(x),\phi(y))=A(x,y)$ for all $x,y\in V$.
