Timeline for Computing the characteristic polynomial without determinants
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 20, 2013 at 21:23 | comment | added | user30035 | Bemao -- I just typed the matrix into my computer and asked it what the char poly was. Whatever did you think I did?? | |
| Feb 17, 2013 at 22:09 | comment | added | Bemao | @quid - Sorry. Yes, there is clearly a way, you're right. I suppose I was looking for something computationally easier than matrix multiplication and then looking for a dependence relation (which is the method I presume wccanard used to find that eqn). I was hoping there would be some way of using the regular structure of the matrix to quickly derive such a relation. | |
| Feb 17, 2013 at 21:52 | comment | added | user9072 | @Bemao: what do you mean by if there is some way to derive them. You can just do matrix-multiplication. So, certainly, there is a way. | |
| Feb 17, 2013 at 18:10 | comment | added | Bemao | @wccanard, what do you mean by "spot"? Is there someway to derive the coeffs for the Ai just from the entries of the matrix? @Yemon Choi, it's difficult to explain. It has to do with the way the truth tables of certain boolean functions split. They each produce matrices of the above form of size 2^n | |
| Feb 17, 2013 at 12:08 | comment | added | user6976 | $\left( {z}^{2}-2\,z-4 \right) \left( {z}^{6}+8\,{z}^{3}-64 \right)$ | |
| Feb 16, 2013 at 0:53 | comment | added | Yemon Choi | In view of Qiaochu's observation: where did this matrix arise? | |
| Feb 15, 2013 at 23:00 | comment | added | user30035 | You could spot that $A^8 - 2A^7 - 4A^6 + 8A^5 - 16A^4 - 32A^3 - 64A^2 + 128A + 256=0$ and then check that $1,A,A^2,\ldots,A^7$ are linearly independent, and hence that this must be the char poly. | |
| Feb 15, 2013 at 22:11 | comment | added | Peter Mueller | Or you could compute the traces of $A$, $A^2$, ..., $A^8$ to obtain the power sums of the eigenvalues, and then use the Newton formula to express the elementary symmetric functions of the eigenvalues (which up to signs are the coefficients of the characteristic polynomial) in terms of the power sums. | |
| Feb 15, 2013 at 22:04 | comment | added | Gerry Myerson | You could calculate $A^2,A^3,\dots,A^8$ and then find a linear dependence relation among $I,A,\dots,A^8$. | |
| Feb 15, 2013 at 21:49 | comment | added | Qiaochu Yuan | This matrix looks structured enough that it should be possible to write down all of the eigenvectors explicitly by some inspired guessing. | |
| Feb 15, 2013 at 21:20 | history | asked | Bemao | CC BY-SA 3.0 |