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?
Best Answer
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. $\square$