31–40 of 64 posts
Earlier quoted context omitted.
It'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.
My brain also automatically added sum and also haven't noticed anything, but I have a math degree, maybe it's just assuming things :-)
The title words remind of an unrelated fact, Gershgorin Disks: The eigenvalues of any N x N matrix, A, are contained in the union of N discs in the complex plane. The center of the i_th disc is the i_th diagonal element of A. The radius of the i_th disc is the absolute values of the off-diagonal elements in the i_th row. https://blogs.sas.com/content/iml/2019/05/22/gershgorin-disc... It's rather remarkable, unexpecte…
What was the noise cancelling project? How did you use this fact to cancel noise?
A common noise cancellation technique is to throw away small eigenvalues, as in PCA. This result relates eigenvalues to the structure of the matrix, so might be helpful for reducing ev's without bothering with diagonalization?
[Edit] This would presumably involve just zeroing out the rows with small diagonal elements and small-ish off-diagonal norm... Center the eigenvalue estimate disk at zero, and then zero out the rest of the row to make the estimate exact.
Earlier quoted context omitted.
Matrix theory by Franklin is a great, affordable book containing many interesting results such as this — can highly recommend for those interested in linear algebra. https://www.amazon.com/Matrix-Theory-Dover-Books-Mathematics...
If I have an atrophied high school level understanding of linear algebra, will I get anything out of that book?
The title words remind of an unrelated fact, Gershgorin Disks: The eigenvalues of any N x N matrix, A, are contained in the union of N discs in the complex plane. The center of the i_th disc is the i_th diagonal element of A. The radius of the i_th disc is the absolute values of the off-diagonal elements in the i_th row. https://blogs.sas.com/content/iml/2019/05/22/gershgorin-disc... It's rather remarkable, unexpecte…
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.
The title words remind of an unrelated fact, Gershgorin Disks: The eigenvalues of any N x N matrix, A, are contained in the union of N discs in the complex plane. The center of the i_th disc is the i_th diagonal element of A. The radius of the i_th disc is the absolute values of the off-diagonal elements in the i_th row. https://blogs.sas.com/content/iml/2019/05/22/gershgorin-disc... It's rather remarkable, unexpecte…
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.
The title words remind of an unrelated fact, Gershgorin Disks: The eigenvalues of any N x N matrix, A, are contained in the union of N discs in the complex plane. The center of the i_th disc is the i_th diagonal element of A. The radius of the i_th disc is the absolute values of the off-diagonal elements in the i_th row. https://blogs.sas.com/content/iml/2019/05/22/gershgorin-disc... It's rather remarkable, unexpecte…
Not a proof or anything, of course the proof is on Wikipedia and nice and elegant. Just a thought on the gut feeling.
I agree that it is a very nice result.
Earlier quoted context omitted.
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.
The Euler's identity is also trivial to deduce, but this doesn't diminish its beauty.
Earlier quoted context omitted.
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.
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.
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.