Timeline for Is there a way to simplify the following trace expression?
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Oct 7, 2015 at 22:00 | vote | accept | user34626 | ||
| Oct 7, 2015 at 21:50 | history | edited | user34626 | CC BY-SA 3.0 |
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| Oct 6, 2015 at 19:33 | answer | added | Carlo Beenakker | timeline score: 8 | |
| Oct 6, 2015 at 19:05 | history | edited | user34626 | CC BY-SA 3.0 |
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| Oct 6, 2015 at 18:39 | comment | added | user34626 | @CarloBeenakker There are $N$ zeros between $c_0$ and $c_{M-1}$ in the first row. I guess the confusion comes by thinking that $M-1$ is the index of $c_{M-1}$ in the first row of the matrix. I agree it's confusing, but I cannot think of any other way to write it. The way I'm constructing the Toeplitz matrix is I'm taking an $M$ long vector $[c_1,...,c_{M-1}]$ and shifting it through every row of the matrix while filling the remaining $N$ spots in every row with zeros. | |
| Oct 6, 2015 at 18:29 | comment | added | Carlo Beenakker | after the edit it looks like $C$ has size $(2M-1)\times(2M-1)$; where does $N$ enter? | |
| Oct 6, 2015 at 18:12 | comment | added | user34626 | @CarloBeenakker Sorry this part was not clear, I updated the question. $[c_0,...,c_{M-1}]$ has length $M$ and the rest of rows and columns are filled with $N$ zeros in the matrix. | |
| Oct 6, 2015 at 18:08 | history | edited | user34626 | CC BY-SA 3.0 |
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| Oct 6, 2015 at 17:10 | comment | added | Carlo Beenakker | it would seem that your matrix $C$ has size $(2n+1)\times(2n+1)$ --- is that what you want? if it is, how does this relate to the size $(N+M)\times(N+M)$? | |
| Oct 6, 2015 at 16:21 | review | First posts | |||
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| Oct 6, 2015 at 16:18 | history | asked | user34626 | CC BY-SA 3.0 |