3 ms·

Being easy to prove doesn't make it unremarkable. Lots of theorems, including this one, have straightforward proofs once you are given the exact formulation. Th

by kmm 3y ago

Being easy to prove doesn't make it unremarkable. Lots of theorems, including this one, have straightforward proofs once you are given the exact formulation. The tricky part is coming up with the idea for the theorem itself. I remember being (mildly) shocked when I was taught this in undergrad, it just seemed too good to be true.

stabbles 3y ago
I think that's just how textbooks present it. A fact is stated but you're lacking intuition. If you had played around a bit with Laplacian matrices, like tri-diagonal matrices with stencil [-1, 2, -1], and found that its eigenvalues are within 2 ± 2, and if you also realized that A + τI has the same eigenvalues shifted by τ, then it's a small step to consider that the magnitude of the off-diagonal may have something to do with the spread of eigenvalues. It's likely that Gerschgorin stumbled upon it like this.
- xelxebar 3y ago
  > tri-diagonal matrices with stencil [-1, 2, -1], Just trying to understand these terms. So is a 5x5 tridiagonal matrix with your stencil look like this? 2 -1 0 0 0 -1 2 -1 0 0 0 -1 2 -1 0 0 0 -1 2 -1 0 0 0 -1 2
  - stabbles 3y ago
    Yeah, exactly. Your matrix shows up when you do a 3-point discretization of -u''(x) = λu(x) with u(0) = u(1) = 0, at 5 equidistant (interior) grid points. That's the discrete, 1D version of the problem they're looking at in the paper. See https://en.wikipedia.org/wiki/Compact_stencil#Three_Point_Stencil_Example https://en.wikipedia.org/wiki/Compact_stencil#Three_Point_St...
llmzero 3y ago
Just thinking about some intuition for the result of the theorem: If the off diagonal elements are zero then the diagonal element is an eigenvalue, by continuity of the determinant, if the off diagonal element are small then $det(A-a_{ii}\lambda)$ is almost zero, that is the new eigenvalue is near aii. So it suggests that the off diagonal elements measure how far is aii from being an eigenvalue.

by kmm 3y ago

stabbles 3y ago

I think that's just how textbooks present it. A fact is stated but you're lacking intuition. If you had played around a bit with Laplacian matrices, like tri-diagonal matrices with stencil [-1, 2, -1], and found that its eigenvalues are within 2 ± 2, and if you also realized that A + τI has the same eigenvalues shifted by τ, then it's a small step to consider that the magnitude of the off-diagonal may have something to do with the spread of eigenvalues. It's likely that Gerschgorin stumbled upon it like this.

xelxebar 3y ago
> tri-diagonal matrices with stencil [-1, 2, -1], Just trying to understand these terms. So is a 5x5 tridiagonal matrix with your stencil look like this? 2 -1 0 0 0 -1 2 -1 0 0 0 -1 2 -1 0 0 0 -1 2 -1 0 0 0 -1 2
- stabbles 3y ago
  Yeah, exactly. Your matrix shows up when you do a 3-point discretization of -u''(x) = λu(x) with u(0) = u(1) = 0, at 5 equidistant (interior) grid points. That's the discrete, 1D version of the problem they're looking at in the paper. See https://en.wikipedia.org/wiki/Compact_stencil#Three_Point_Stencil_Example https://en.wikipedia.org/wiki/Compact_stencil#Three_Point_St...

llmzero 3y ago

Just thinking about some intuition for the result of the theorem: If the off diagonal elements are zero then the diagonal element is an eigenvalue, by continuity of the determinant, if the off diagonal element are small then $det(A-a_{ii}\lambda)$ is almost zero, that is the new eigenvalue is near aii. So it suggests that the off diagonal elements measure how far is aii from being an eigenvalue.