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one more backtick block

Glen Whitney 2025-02-19 20:40:47 +00:00
parent 5c4d46cea8
commit f876541d4c
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Gram-matrix-parameterization.md

@ -64,12 +64,12 @@ It follows that
Since the subspace $C$ is transpose-invariant, we also have Since the subspace $C$ is transpose-invariant, we also have
\[ \mathcal{P}(X^\top) = \mathcal{P}(X)^\top. \] \[ \mathcal{P}(X^\top) = \mathcal{P}(X)^\top. \]
We can now see that We can now see that
\[ ```math
\begin{align*} \begin{align*}
d\Delta & = -\mathcal{P}(dA^\top Q A + A^\top Q\,dA) \\ d\Delta & = -\mathcal{P}(dA^\top Q A + A^\top Q\,dA) \\
& = -\big[\mathcal{P}(A^\top Q\,dA)^\top + \mathcal{P}(A^\top Q\,dA)\big]. & = -\big[\mathcal{P}(A^\top Q\,dA)^\top + \mathcal{P}(A^\top Q\,dA)\big].
\end{align*} \end{align*}
\] ```
Plugging this into our formula for $df$, and recalling that $\Delta$ is symmetric, we get Plugging this into our formula for $df$, and recalling that $\Delta$ is symmetric, we get
```math ```math
\begin{align*} \begin{align*}