Hi,
For triangular matrices, it's straightforward to calculate eigenvectors.
You just need triangular solves. See Section 4.4.1 of Heath's Scientific
Computing 2nd Edition.
Isuru
On Tue, Oct 9, 2018 at 11:27 AM Jacob Miner wrote:
> I will show you a representation of the 7x7 form of my
I will show you a representation of the 7x7 form of my matrix, the 10x10
includes a couple additional elements, but has the same overall structure
and layout.
The key point is that the diagonal elements are differences of multiple
values, and each of these values occupies a certain element in
Your matrix is far simpler than I had imagined (you should have
mentioned that it was triangular). I think as Isuru said we can likely
implement a faster method for triangular matrices. The eigenvalues
themselves (the diagonals) are already computed very quickly.
Aaron Meurer
On Tue, Oct 9, 2018
I think I understand, but is there an implementation of this technique that
can actually perform the linear algebra on a symbolic matrix at such
improved compute-time?
On Tuesday, October 9, 2018 at 1:58:04 PM UTC-6, Isuru Fernando wrote:
>
> First k-1 entries of the k th eigenvector for an
Isuru,
I went into Heath's text to get your reference, and it helps layout the
method, but can you please clarify what you meant by 'triangular solves'?
Thank you.
On Tue, Oct 9, 2018, 10:45 Aaron Meurer wrote:
> Your matrix is far simpler than I had imagined (you should have
> mentioned that
First k-1 entries of the k th eigenvector for an upper triangular matrix U
is U[:k-1,:k-1]^-1 @ U[:k-1,k], which is a triangular solve since
U[:k-1,:k-1] is a triangular matrix and it can be done in O(k^2) time.
Isuru
On Tue, Oct 9, 2018 at 1:27 PM Jacob Miner wrote:
> Isuru,
>
> I went into
