Timeline for Existence of a matrix product from its eigenvalues
Current License: CC BY-SA 3.0
18 events
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| Jan 9, 2015 at 16:10 | history | edited | ScienceSnake | CC BY-SA 3.0 |
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| Jan 9, 2015 at 15:16 | history | edited | ScienceSnake | CC BY-SA 3.0 |
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| Jan 9, 2015 at 15:07 | history | edited | ScienceSnake | CC BY-SA 3.0 |
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| Jan 9, 2015 at 15:03 | comment | added | ScienceSnake | I've checked the example above: the eigenvalues of A, B and AB are definitely right. I do however agree with you Shamisen, something doesn't make sense. In fact, there is no reason in general that $\sum_i \lambda^\downarrow(A)_i \lambda^\uparrow(B)_i = \sum_i \lambda^\downarrow(A)_i \lambda^\downarrow(B)_i$ so it's not even necessarily true that $\lambda^\downarrow(A) \cdot \lambda^\uparrow(B)\prec \lambda^\downarrow(A) \cdot \lambda^\downarrow(B)$. Ive looked through the textbook that first states this (Bhatia) and the only explanation I can find is that he means $\prec_w$ instead of $\prec$. | |
| Jan 9, 2015 at 2:41 | comment | added | Tadashi | @Ben I don't think this example is correct, since $\lambda^\downarrow(A) \cdot \lambda^\uparrow(B) = (2,3)$,$\lambda(AB) = (4, 1.5)$ and hence $2+3 \ne 4 + 1.5$. | |
| Jan 9, 2015 at 2:38 | comment | added | Tadashi | @StevenGubkin $\prec$ means the majorization preorder and $x \cdot y := (x_1 y_1, \ldots, x_n y_n)$. | |
| Jan 8, 2015 at 20:15 | comment | added | Steven Gubkin | I guess I am maybe more confused than I thought. If $\lambda(A)$ is only a set (not an ordered set), what does $\prec$ mean here? | |
| Jan 8, 2015 at 20:12 | comment | added | Steven Gubkin | I am probably only proving how stupid I am at this point, but how does taking $A$ and $B$ to be diagonal matrices above not invalidate the theorem? | |
| Jan 8, 2015 at 20:09 | comment | added | Steven Gubkin | Thanks! This clarifies things immensely. Would you mind adding this example to the question? | |
| Jan 8, 2015 at 20:02 | comment | added | ScienceSnake | Let's take an explicit example. Let $\lambda(A)=(2,1)$, $\lambda(B)=(3,1)$, and $\lambda(AB)=(4,1.5)$. It's easy to check that these numbers satisfy the inequality in the question. Clearly, taking A and B to be diagonal matrices, however, does not work. In this case, a solution is to take $$ A = \left( \begin{array}{cc} \frac{5}{4} & \frac{\sqrt{3}}{4} \\ \frac{\sqrt{3}}{4} & \frac{7}{4} \\ \end{array} \right)$$ and $$B = \left( \begin{array}{cc} 3 & 0 \\ 0 & 1 \\ \end{array} \right)$$. | |
| Jan 8, 2015 at 19:47 | review | Close votes | |||
| Jan 8, 2015 at 20:44 | |||||
| Jan 8, 2015 at 19:44 | comment | added | Steven Gubkin | If I understand everything correctly, just define $A$ and $B$ to be diagonal matrices, where the eigenvalues are placed in descending order. Can you confirm this is what you are asking about? If so it is really off topic for this site. | |
| Jan 8, 2015 at 19:34 | comment | added | ScienceSnake | Reworded the question to, hopefully, be clearer. | |
| Jan 8, 2015 at 19:32 | history | edited | ScienceSnake | CC BY-SA 3.0 |
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| Jan 8, 2015 at 19:21 | comment | added | Steven Gubkin | How about $A$ is the identity matrix, and $B$ is the diagonal matrix with the given eigenvalues? Can you make your question a bit more precise and generally understandable? | |
| Jan 8, 2015 at 18:35 | history | edited | ScienceSnake | CC BY-SA 3.0 |
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| Jan 8, 2015 at 16:45 | review | First posts | |||
| Jan 8, 2015 at 17:09 | |||||
| Jan 8, 2015 at 16:43 | history | asked | ScienceSnake | CC BY-SA 3.0 |