Proving symplectic identity

Question

Let $\Lambda$ be a skew-symmetric matrix and $Q$ a symmetric matrix. Let $\text{Id}$ be the identity matrix and $h > 0$ a real number. I am trying to prove the following identity: $$ (\text{Id} + \frac{h}{2} \Lambda Q) \Lambda (\text{Id} + \frac{h}{2} \Lambda Q)^\top = (\text{Id} - \frac{h}{2} \Lambda Q) \Lambda (\text{Id} - \frac{h}{2} \Lambda Q)^\top $$ Here is my attempt at this where I use the facts that $Q^\top=Q$ and $\Lambda^\top = -\Lambda$. We have, \begin{align} (\text{Id} + \frac{h}{2} \Lambda Q) \Lambda (\text{Id} + \frac{h}{2} \Lambda Q)^\top &= (\text{Id} - \frac{h}{2} \Lambda^\top Q) \Lambda (\text{Id} - \frac{h}{2} \Lambda^\top Q)^\top \\ &= (\text{Id} - \frac{h}{2} \Lambda^\top Q) \Lambda (\text{Id} - \frac{h}{2} Q\Lambda) \\ &= (\text{Id} - \frac{h}{2} Q\Lambda )^\top \Lambda (\text{Id} - \frac{h}{2} Q\Lambda) \end{align} This looks almost correct but I've got the transpose on the wrong side. Does anyone see how I can manipulate this to the correct form?

Answer 1

Note that $(\text{Id} + \frac{h}{2} \Lambda Q) \Lambda = \Lambda(\text{Id}+\frac{h}{2} Q\Lambda)$ and $(\text{Id} + \frac{h}{2} \Lambda Q)^\top = \text{Id}-\frac{h}{2}Q\Lambda$.

Thus \begin{align} (\text{Id} + \frac{h}{2} \Lambda Q) \Lambda (\text{Id} + \frac{h}{2} \Lambda Q)^\top &= \Lambda(\text{Id}+\frac{h}{2} Q\Lambda)(\text{Id}-\frac{h}{2}Q\Lambda) \\ &= \Lambda(\text{Id} - \frac{h^2}{4}Q\Lambda Q\Lambda) \end{align}

The same argument for $-\Lambda$ (which is also skew-symmetric) yields \begin{align} (\text{Id} - \frac{h}{2} \Lambda Q) \Lambda(\text{Id} - \frac{h}{2} \Lambda Q)^\top &= -(\text{Id} - \frac{h}{2} \Lambda Q) (-\Lambda)(\text{Id} - \frac{h}{2} \Lambda Q)^\top \\ &= -(-\Lambda)\left(\text{Id} - \frac{h^2}{4}Q(-\Lambda) Q(-\Lambda)\right) \\ &= \Lambda(\text{Id} - \frac{h^2}{4}Q\Lambda Q\Lambda) \\ &= (\text{Id} + \frac{h}{2} \Lambda Q) \Lambda (\text{Id} + \frac{h}{2} \Lambda Q)^\top \end{align}

as desired.