4 ms·
There's a geometric solution as well! Since M is invertible and symmetric. It acts on R^N by independently scaling the coordinates of an orthogonal basis. Eg,
by WoahNoun 4y ago
There's a geometric solution as well!
Since M is invertible and symmetric. It acts on R^N by independently scaling the coordinates of an orthogonal basis. Eg, M = O^T*A*O where O is orthogonal and A is a diagonal matrix. Let sqrt(A) be the diagonal matrix with entries equal to the square root of the corresponding entry in A. Then K = O^T*sqrt(A)*O is symmetric and K^2 = (O^T*sqrt(A)*O)*(O^T*sqrt(A)*O). Since O is orthogonal O^T*O = I, and K^2 = O^T*A*O = M.
- ogogmad 4y agoThat doesn't work. It's complex symmetric, but not real symmetric or Hermitian, so you can't use the spectral theorem. This statement here: It acts on R^N by independently scaling the coordinates of an orthogonal basis is precisely the spectral theorem, which you can't use here. [edit] Also, the corresponding claim for Hermitian (and real symmetric) matrices isn't true either. If an eigenvalue of a Hermitian matrix H is negative, then no square root of it can be Hermitian (or real symmetric).
- WoahNoun 4y agoAh I missed the complex part. Clearly a complex matrix doesn't act on R^N by scaling the coordinates.