Timeline for Similarity of two matrices
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 26, 2021 at 17:43 | comment | added | Antoine Labelle | The matrix for a fixed $\lambda$ is diagonalizable, but there is no base that simultaneously (block-)diagonalize all $A(\lambda)$ | |
| May 26, 2021 at 17:31 | comment | added | Pritam Bemis | actually, I do have a stupid question. I mean after all, the matrix is diagonalizable since we know what its distinct eigenvalues are, no? So how is it possible that you show it cannot be block-diagonalized, which seems like a weaker question? | |
| May 25, 2021 at 19:51 | comment | added | Antoine Labelle | Let us continue this discussion in chat. | |
| May 25, 2021 at 19:32 | comment | added | Pritam Bemis | I see that for $\phi=\pi/2$ the matrix $A$ is skew-symmetric. How do you conclude from this that the product of such matrices has complex conjugate pairs as eigenvalues? (the only question is really why one complex conjugate pair exists, I understand how to show that there are two then in that case) | |
| May 25, 2021 at 18:57 | comment | added | Antoine Labelle | Interesting, I wonder what explains this phenomenon this time (in the case $\phi=\pi/2$ it follows directly from skew-hermiticity but that doesn't work for general $\phi$) | |
| May 25, 2021 at 18:24 | comment | added | Pritam Bemis | sorry, degenerate or complex conjugate pairs. I corrected it after I wrote it the first time. But nevertheless, what I mean is, it is just the same as with the $\mu$ in the other post. | |
| May 25, 2021 at 18:23 | comment | added | Antoine Labelle | The matrices do not have degenerate eigenvalues in general, actually for $\phi=\pi/2$ the matrix $A$ has 4 distinct eigenvalues for any $\lambda$. | |
| May 25, 2021 at 18:19 | comment | added | Pritam Bemis | Sorry for being vague. Call the matrix I defined $A(\lambda,\alpha)$. Let $\lambda_i$ be a sequence of real numbers, then $\prod_{i=1}^n A(\lambda_i,\alpha)$ has two double degenerate/complex conjugate eigenvalues. This is precisely what happened also in this other post. But apparently, this time it is not because of some block-diagonal form? | |
| May 25, 2021 at 18:16 | comment | added | Antoine Labelle | What do you mean by "the degenerate eigenvalues are stable under taking products of such matrices for different $\lambda$"? | |
| May 25, 2021 at 18:13 | vote | accept | Pritam Bemis | ||
| May 25, 2021 at 18:08 | comment | added | Pritam Bemis | I should add that actually I only asked this question trying to generalize what you did in the other post, but somehow it does not work, as you proved yourself. However, the result still seems to apply. | |
| May 25, 2021 at 18:03 | comment | added | Pritam Bemis | interesting, let me go in detail through your answer. but somehow i am surprised about this negative answer, as I saw you answered another similar question, where you showed that eigenvalues always come in pairs, just as here and obtained a block reduction. In fact, just as in the other answer, I now also checked that the degenerate eigenvalues are stable under taking products of such matrices for different $\lambda$ but fixed $\varphi$. Strange, don't you think? | |
| May 25, 2021 at 18:00 | history | answered | Antoine Labelle | CC BY-SA 4.0 |