6 ms·

It's not that remarkable at all? The proof requires the definition and triangle inequality, that's all? Given Ax=λx, take i for which |xᵢ| is largest. Look at

It's not that remarkable at all? The proof requires the definition and triangle inequality, that's all? Given Ax=λx, take i for which |xᵢ| is largest. Look at the i'th equation: sum aᵢⱼxⱼ = λxᵢ, move the aᵢᵢxᵢ term to the rhs, take absolute values, divide by |xᵢ|, apply triangle inequality, and you have |aᵢᵢ - λ| ≤ sum |aᵢⱼ| over j≠i. So for every eigenvalue you can find such a disc. That's by column, for row use Aᵀ.

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.
nine_k 3y ago
The Euler's identity is also trivial to deduce, but this doesn't diminish its beauty.
- xanderlewis 3y ago
  The Yoneda lemma is another great example of (once you've got the right setup) trivial but beautiful
moralestapia 3y ago
Huh? Math 101, simple is better.
jonahx 3y ago
removed
- tsimionescu 3y ago
  They are saying that the Greshgorin theorem that the OP is talking about is simple to prove, not the Polya conjecture that took 70 years from the article.

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.

nine_k 3y ago

The Euler's identity is also trivial to deduce, but this doesn't diminish its beauty.

xanderlewis 3y ago
The Yoneda lemma is another great example of (once you've got the right setup) trivial but beautiful

moralestapia 3y ago

Huh? Math 101, simple is better.

jonahx 3y ago

removed

tsimionescu 3y ago
They are saying that the Greshgorin theorem that the OP is talking about is simple to prove, not the Polya conjecture that took 70 years from the article.