5 ms·
TL;DR: This lets one calculate eigenvectors directly from the eigenvalues of a matrix and its submatrices, Element wise: v(i,j)=√∏δ(k,i)(λi(A) − λk(A))/δ(k,n)
by singularity2001 7y ago
TL;DR: This lets one calculate eigenvectors directly from the eigenvalues of a matrix and its submatrices,
Element wise:
v(i,j)=√∏δ(k,i)(λi(A) − λk(A))/δ(k,n)(λi(A) − λk(Mj))
δ(x,y)(f):= ±1 if x=y else f
- osipov 7y agoDoes this yield a more parallelizable algorithm compared to what we had previously?
- lostmsu 7y agoBTW, I discovered a cool service recently: HTML pastebin, that supports math. E.g. https://pste.eu/p/3B68.html https://pste.eu/p/3B68.html
- plus 7y agoWhere does sign on v(i,j) come from? The arxiv paper says that it is possible to extract the phase factor, but it does not say how.
- jhallenworld 7y agoSo the textbook way is: 1. Roots of characteristic equation det(A-λI)=0 give the eigenvalues. 2. Solve (A-λI)x = 0 for each eigenvalue to find each eigenvector x. It seems very plausible that step 2 can be reduced to an equation.