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V(:,:,ii) = inv(K)*U*diag(sqrt(gCommd.*omega(:, hklIdxMEM(ii)))); %#ok<MINV> |
The calculation calculates the transformation matrix $V$ as $V = K^{-1} U E^{1/2}$. Calculating the inverse of a matrix to use in a calculation is usually not recommended as it is quite unstable (MATLAB docs find for a 500x500 matrix the error is 10^9 larger!) compared to calculating $V$ as the solution to the linear system $KV = UE^{1/2}.$
The solve would also be more efficient - inv(K) works by solving the linear system $K X = I$ for $X$ (so we get that $X$ is $K^{-1}$). So if we calculate inv(K) purely to calculate V, we are doing a linear system solve and then two unnecessary matrix multiplications.
spinw/swfiles/@spinw/spinwave.m
Line 1013 in dc06a4b
The calculation calculates the transformation matrix$V$ as $V = K^{-1} U E^{1/2}$ . Calculating the inverse of a matrix to use in a calculation is usually not recommended as it is quite unstable (MATLAB docs find for a 500x500 matrix the error is 10^9 larger!) compared to calculating $V$ as the solution to the linear system $KV = UE^{1/2}.$
The solve would also be more efficient -$K X = I$ for $X$ (so we get that $X$ is $K^{-1}$ ). So if we calculate
inv(K)works by solving the linear systeminv(K)purely to calculateV, we are doing a linear system solve and then two unnecessary matrix multiplications.