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Derivative of a matrix inverse The derivative of a matrix inverse $\boldsymbol{A}^{-1}$ with respect to $\boldsymbol{A}$ is a 4th order tensor. Since $\dfrac{\partial}{\partial \boldsymbol{A}} \left( \boldsymbol{A}^{-1} \boldsymbol{A} \right) = 0$, $\dfrac{\partial \boldsymbol{A}^{-1}}{\partial \boldsymbol{A}} = - \boldsymbol{A}^{-2}$, but deriving its counterpart based on Einstein notation is not so obvious. Owing to, $$ 0 = \frac{\partial \delta_{km}}{\partial A_{ij}} = \frac{\partial A_{kl}^{-1}}{\partial A_{ij}} A_{lm}+A_{kl}^{-1} \frac{\partial A_{lm}}{\partial A_{ij}} $$and $$ \frac{\partial A_{kn}^{-1}}{\partial A_{ij}} = \frac{\partial A_{kl}^{-1}}{\partial A_{ij}} A_{lm} A_{mn}^{-1} = - A_{kl}^{-1} \frac{\partial A_{lm}}{\partial A_{ij}} A_{mn}^{-1}, $$where ...

The Orthogonal Decomposition Theorem Let $V$ be a Hilbert space and $M\subset V$ a closed subspace of $V$. Then (i) $M^\bot$ is a closed subspace of $V$. (ii) $V$ can be represented as the direct sum of $M$ and its orthogonal complement $M^\bot$ $$ V=M\oplus M^\bot $$i.e., every vector $v\in V$ can be uniquely decomposed into two orthogonal vectors $\boldsymbol m$, $\boldsymbol n$, s.t. $$ \boldsymbol v=\boldsymbol m+\boldsymbol n, \boldsymbol m\in M,\boldsymbol n\in M^\bot $$COROLLARY Let $V$ be a Hilbert space and $M$ a vector subspace of $V$ . The following conditions are equivalent to each other (i) $M$ is closed. (ii) $(M^\bot)^\bot = M$. ...

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