linear algebra - Should the formula for the inverse of a 2x2 matrix be obvious?

matrix evaluation – Gradient Descent for Markov Dynamics Answer

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matrix evaluation – Gradient Descent for Markov Dynamics

Note that $A$ solely seems within the mixture $M=A-BC$, so the by-product with respect to $A$ equals the by-product with respect to $M$; The duty $f(A)$ is given by
$$f(A)=||(A – BC)^Nv – w||_2^2=(M^Nv)^T M^Nv+w^Tw-2w^TM^Nv.$$
For a unostentatious illustration, let me first deem a scalar perturbation, $f(A+epsilon I)=f(A)+epsilon df/dA$, with by-product
$$frac{df}{dA} =2N(M^Nv-w)^TM^{N-1}v.$$

Next deem the by-product with respect to a given matrix component $A_{ij}$ of $A$. The expressions are extra prolonged, mainly every matrix $M$ offers a several time period so we’ve a sum $sum_{okay=1}^N$ as a substitute of the issue $N$:
$$frac{partial f}{partial A_{ij}}= 2sum_{okay=1}^Nsum_{p,q=1}^n (M^Nv-w)_q (M^{k-1})_{qi} (M^{N-k})_{jp} v_p.$$
Note that the scalar perturbation is the vestige of the matrix of elementwise perturbations, $df/dA=sum_{i=1}^n partial f/partial A_{ii}$.


As celebrated within the observation, my outcome for the elementwise by-product doesn’t appear to conform with the one from a web-based calculator. Let me bridle a unostentatious illustration, $N=2$, $w=0$, $v_p=delta_{p1}$. Then
$$f=sum_{p,q,r}M_{pq}M_{q1}M_{pr}M_{r1}.$$
Direct analysis of the by-product with respect to $M_{22}$ offers
$$frac{partial f}{partial M_{22}}=2M_{21}(M^2)_{21},$$
in settlement with the common method above. The on-line calculator would give 0.

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