Timeline for How to find the eigenvectors and eigenvalues of a hermitian operator?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 24, 2023 at 10:51 | vote | accept | zizaaooo | ||
| Dec 24, 2023 at 10:26 | comment | added | Tristan Nemoz♦ | @zizaaooo Oops, you're right! Its determinant isn't nil, so $M-\lambda I$ is invertible indeed, sorry for that! | |
| Dec 24, 2023 at 10:15 | comment | added | zizaaooo | Do you mean that $M-\lambda$ is invertible because the only solution is $x=0$ if the $\lambda$ is not an eigenvalue of $M$? Because you wrote in the first line of your comment that $M-\lambda$ isn't invertible. I might be repeating what you just said but I want to make sure that I understood your answer. | |
| Dec 24, 2023 at 9:27 | comment | added | Tristan Nemoz♦ | @zizaaooo In this step, our goal is to show that if $\lambda$ is not an eigenvalue of $M$, then $M-\lambda I$ isn't invertible. In order to do this, we use the second part of Fact 2. : we show that the only $x$ satisfying $(M-\lambda I)x=0$ is $x=0$. | |
| Dec 24, 2023 at 8:49 | comment | added | zizaaooo | Thanks a lot, but I why did you make a point about $x$ being 0 if $\lambda$ is not an eigenvalue of $M$? | |
| Dec 23, 2023 at 10:11 | history | edited | Tristan Nemoz♦ | CC BY-SA 4.0 |
deleted 46 characters in body
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| Dec 23, 2023 at 5:14 | comment | added | FDGod | As for learning linear algebra, I would highly recommend Greg Sanderson's Essence of Linear Algebra series to get an excellent intuitive understanding. If you are looking for something slightly advanced, then I would recommend Gilbert Strang's Linear Algebra Course at MIT. | |
| Dec 23, 2023 at 0:14 | history | answered | Tristan Nemoz♦ | CC BY-SA 4.0 |