Timeline for How to show a certain determinant is non-zero
Current License: CC BY-SA 3.0
14 events
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| Mar 28, 2013 at 18:34 | history | edited | Suvrit | CC BY-SA 3.0 |
added comment thanks to Todd's counterexamples.
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| Mar 28, 2013 at 18:27 | comment | added | Suvrit | Thanks Todd! Nice counterexample. Clearly, what I've written is not sufficient to conclude the claim. I'm still thinking, however, that a variation of my outline in the answer could somehow be made to do the job---though a further argument does not immed. come to my mind. (just another junk remark though: we know that in our case $AB^T$ has strictly positive entries, so arbitrary cancellations cannot happen, but this again is not going to suffice!). If I get a chance, I'll try to rescue my proof attempt, otherwise, I hope someone else on MO takes the time to rescue it :-) | |
| Mar 28, 2013 at 16:41 | comment | added | Todd Trimble | Hate to be such a spoil-sport (and I'd welcome a variation on your idea that gets the job done), but let me give a simple example of the type of thing I think needs to be addressed. Take $n=1$, and suppose $A$ is the $1 \times \infty$ matrix $(1, 1, 1, \ldots)$, and $B$ is the $1 \times \infty$ matrix $(1, -1/2, -1/4, -1/8, \ldots)$. Then every set of $n$ columns of $A$ and of $B$ is linearly independent. But $A B^T = 0$. | |
| Mar 28, 2013 at 0:17 | comment | added | Suvrit | except perhaps if one could use the fact that every set of $n$ columns of $L$ (and $X$) is linearly ID---maybe that could rescue the proof... | |
| Mar 28, 2013 at 0:15 | comment | added | Suvrit | You are right Todd; I forgot that $L$ and $X$ are $n \times \infty$, so I cannot just conclude that $LX^T$ has the desired rank. And to think that I complained that my students have forgotten linear algebra ;-) --- now it seems that without a new idea, the above proof is not going further.... | |
| Mar 27, 2013 at 23:31 | comment | added | Todd Trimble | I'm sorry, Suvrit, but I still don't follow. I agree that both $L$ and $X$ have rank $n$. But the product of two rank $n$ matrices need not be of rank $n$. For example, it is easy to give transformations $A: \mathbb{R}^n \to \mathbb{R}^{2n}$ and $B: \mathbb{R}^{2n} \to \mathbb{R}^n$, both of rank $n$, but where the image of $A$ lies in the kernel of $B$. | |
| Mar 27, 2013 at 23:14 | history | edited | Suvrit | CC BY-SA 3.0 |
please see comments for additional clarification.
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| Mar 27, 2013 at 21:53 | comment | added | Suvrit | But Todd, for each $k \ge n$, the matrix $L_k$ has full rank, as does $X_k$; also, $\lim_{k \to \infty} L_k = L$ has full rank (already the first $n$ columns of $L$ are LI); similarly, $\lim_{k \to \infty} X_k=X$ has full rank. Thus, in particular, $LX^T$ has full rank. Additionally, $\lim_{k\to \infty} L_kX_k^T = \lim_k L_k \lim_k X_k^T = LX^T = M$ is also well-defined (so that we know how to multiply these infinite matrices). | |
| Mar 27, 2013 at 21:41 | history | edited | Suvrit | CC BY-SA 3.0 |
added the "caveat"
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| Mar 27, 2013 at 19:55 | comment | added | Todd Trimble | Thanks. I had already understood what you put in the note; my worry (which I think you understood) is elsewhere. | |
| Mar 27, 2013 at 19:35 | comment | added | Suvrit | @Todd: I added a short note of clarification; but I see your worries if we view $M = \lim_{k \to \infty) L_kX_k^T$, because rank is not closed. I just viewed $(LX^T)_{ij}$ as a power series (which converges everywhere, and thus is ok. But maybe this needs to be made slightly more rigorous.... | |
| Mar 27, 2013 at 19:31 | history | edited | Suvrit | CC BY-SA 3.0 |
fixed typo, added short note
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| Mar 27, 2013 at 18:54 | comment | added | Todd Trimble | Sorry, I'm having trouble understanding your argument. It looks like all that is being said is that the matrix $M$ of the OP is a limit of matrices of full rank $n$. (I don't know how to interpret the product $L X^T$, which involves infinite sums, unless I interpret it as a limit.) But of course, that doesn't prove $M$ itself is of full rank. | |
| Mar 26, 2013 at 21:56 | history | answered | Suvrit | CC BY-SA 3.0 |