Hello, lovetski, you wrote: L> Hello, kfmn, you wrote: K>> All greetings! K>> I again with the algebra... K>> It is required to solve system of equations: ASA * = B, concerning unknown persons A and S (the asterisk on top designates complex conjugation). K>> Thus it is known that K>> K>> A - symmetric, and on a diagonal it has units K>> S - diagonal, real K>> B - self-interfaced K>> K>> Total, if dimensionality of matrixes NxN, turns out N * (N+1)/2 the equations (considering self-conjugacy B) and as much unknown persons: N * (N-1)/2 elements A and N diagonal elements S. L> And it did not try to use? L> Normal matrices on page L> https://en.wikipedia.org/wiki/Eigendeco
f_a_matrix L> If hardly to twist, that, can be and what is necessary turns out? Yes, with eigen-decomposition I pottered. As the initial matrix B self-interfaced and even nonnegatively defined all its own values are real and nonnegative, i.e. Expansion under latent vectors looks like B=VCV-1, and V-1=V H, i.e. reverse - diagonal the C coincides with nonnegative elements with Hermite-interfaced, and. Then if to enter A = VC 1/2V T, A * =V *C 1/2V H and as a result AA * =B. But how now to make, that at A on a diagonal there were units and in product there was a diagonal multiplicand, I will not think If to divide in any way it on the right or at the left into a scalar matrix, its symmetry disappears, and it for me is critical.