Timeline for Prove that $(v^Tx)^2-(u^Tx)^2 < 1-(u^Tv)^2$ for any unit vectors $u$, $v$, $x$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 28, 2022 at 1:19 | vote | accept | Dan Feldman | ||
| Jan 30, 2022 at 0:56 | comment | added | GH from MO | @QueenMath I think what went wrong is that you forgot the square in the end. The square of the largest eigenvalue is not the maximum of $|(x^Tu)^2-(v^Tx)^2|$, but the maximum of $|(x^Tu)^2-(v^Tx)^2|^2$. | |
| Jan 30, 2022 at 0:39 | comment | added | Dan Feldman | Apologies for the confusion, but some clarifications are still needed: 1) The long proof is from the famous cited paper in the question. 2) I was always referring to L_2 and not Frobenius norm of a matrix, including when writing $||uu^T-vv^T||^2_2$ above. Note that $uu^T$ is a matrix, not a vector. 3) In the second edit I explained the connection between the title and $||uu^T=vv^T||_2$. I assume it is correct, unlike the title, but still do not understand where is the bug in the reduction... | |
| Jan 30, 2022 at 0:28 | comment | added | GH from MO | @QueenMath Your question (as in the title) was about an inequality that is false. This means that whatever complicated proof you had, it was incorrect. Now your identity has little to do with your inequality. Your confusion seems to be two-fold. First, you seem to confuse the Frobenius norm and the $L^2$-norm of a matrix. In your previous remark, you talk about the $L^2$-norm of $A$. The square of this norm is the maximum of $\|A x\|^2=x^TA^TAx$ over the unit vectors $x$, and it has little to do with the maximum of $(v^Tx)^2-(u^Tx)^2$ over the unit vectors $x$. | |
| Jan 30, 2022 at 0:23 | comment | added | Dan Feldman | Thanks again, but I did not meant to ask a new question. I was referring to the conjecture $||uu^T-vv^T||^2_2=1-(u^Tv)^2$ in the question. | |
| Jan 30, 2022 at 0:17 | comment | added | Dan Feldman | ||A||_2 in my question is the largest eigenvalue of a matrix, $\max_x ||Ax||_2/||x||$ where $A=uu^T-vv^T$ is a matrix. Not the Frobenius norm (sum of squared entries) of $A$. | |
| Jan 30, 2022 at 0:15 | comment | added | GH from MO | @QueenMath Please ask separate questions in a separate post. I answered your question. Your conjecture is false. I fixed it and generalized it to an arbitrary number of vectors. | |
| Jan 30, 2022 at 0:14 | comment | added | Nathaniel Johnston | @QueenMath - Do you mean $\|uu^T - vv^T\|_2^2 = 2 - 2(u^Tv)^2$? That equality follows immediately from the definition of the Frobenius norm (by subscript $2$, do you mean the Frobenius norm, aka the Schatten 2-norm?). | |
| Jan 30, 2022 at 0:07 | comment | added | Dan Feldman | Thanks! Do you agree with the original equality $||uu^T-vv^T||^2_2=1-(u^Tv)^2$? If yes, can we use a similar technique to prove it? | |
| Jan 30, 2022 at 0:06 | history | edited | GH from MO | CC BY-SA 4.0 |
added 6 characters in body
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| Jan 29, 2022 at 23:58 | history | answered | GH from MO | CC BY-SA 4.0 |