3 ms·
Sounds very cool. One thing I didn’t understand though: > The radius of the i_th disc is the absolute values How can a radius of a single disc (i.e. a single
by frutiger 3y ago
Sounds very cool. One thing I didn’t understand though:
> The radius of the i_th disc is the absolute values
How can a radius of a single disc (i.e. a single value) correspond to multiple values?
- reactordev 3y agoI’m assuming, but could be wrong as my matrix math ended with GameDev, that the radius of the disk = the absolute values of the other things. +\- depending on which side of the disk they fall? That line got me as well.
- Sharlin 3y agoIt's missing the word "sum". Which apparently my brain auto-deduced for me because I didn't notice anything off in the sentence on the first reading.
- ximeng 3y agoIf you look at the link there’s a formula showing it’s the sum.
- abetusk 3y agoSpecifically: r_i = \sum_{i \ne j} |A_{i j}|
- tomrod 3y agoSum the absolute values of a row for all the off diagonal elements
- Infinity315 3y agoAn intuitive explaination is imagine a nearly diagonal matrix where the values along the diagonal are much larger than values on the off diagonal. We know the eigenvalues of a diagonal matrix is simply the values on the diagonal, so for nearly diagonal matrices you can be pretty sure that the true eigenvalues are going to be pretty close to those diagonal entries, but a natural question to ask is how far we'd deviate from those diagonal entries. The answer to the above question is Gerschgorin disks and it's closely related cousin Brauer's Oval of Cassini. For matrices with real eigenvalues it's moreso along the real number line, only for cases where the eigenvalues are imaginary do we imagine disks.