/sci/ - Science & Math » Thread #14116758

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Linear algebra

Anonymous Tue 18 Jan 00:56:16 2022 No.14116758 View Reply Original Report

Quoted By: >>14117194 >>14117223 >>14118700

How would /sci/ solve that question ?
I proved the theorem, but i have no idea how to find the answer to the question.

Anonymous

Anonymous Tue 18 Jan 2022 01:10:09 No.14116798 Report

Quoted By: >>14116830

what "does non-negative solution Xi" mean in the context of the question? Xi is a vector...

anyway, it seems that to be able to partition a matrix in this way, you'd need to ensure all the submatrices are invertible, which means if you for example a M(3,4) matrix no partition could be made because you'd need n=some multiple of m to have a chance at invertibility

Anonymous

Anonymous Tue 18 Jan 2022 01:22:53 No.14116830 Report

Quoted By:

>>14116798
it means that every coordinate of Xi is non negative.

Anonymous

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Anonymous Tue 18 Jan 2022 01:43:28 No.14116921 Report

Quoted By:

I can give two examples (although these might not be the best examples, considering that in these cases, the given partitions work on any b...)

Anonymous

View Same Google iqdb SauceNAO exemple2.png, 13KiB, 395x221

Anonymous Tue 18 Jan 2022 01:46:20 No.14116933 Report

Quoted By:

Example n°2.

Anonymous

Anonymous Tue 18 Jan 2022 02:45:25 No.14117194 Report

Quoted By: >>14118299

>>14116758
I can't read.

More specifically though, what the fuck does M_i^(b) mean? And what does M_i^(b)X_i mean?

Anonymous

Anonymous Tue 18 Jan 2022 02:50:46 No.14117223 Report

Quoted By: >>14117308 >>14118281

>>14116758
can you give me the book or pdf where this written? i want to save the problem for later

Anonymous

Anonymous Tue 18 Jan 2022 03:27:37 No.14117308 Report

Quoted By:

>>14117223
linear algebra prasolov

Anonymous

Anonymous Tue 18 Jan 2022 10:55:12 No.14118281 Report

Quoted By:

>>14117223
I wrote that myself

Anonymous

Anonymous Tue 18 Jan 2022 11:01:39 No.14118299 Report

Quoted By:

>>14117194
M_i^{(b)} is a subset of the columns of M, and that subset is dependent on the value of b. Meaning that i give you the vector b, and AFTER I've given you that information you can choose the value of M_i^{(b).
The question is can you choose
two subset M_1 and M_2 that are guaranteed to work even before I tell you the value of b ?

M_i^(b)X_i is the product of the submatrix associated with M_i^{b} by some unknown vector.

Anon

Anon Tue 18 Jan 2022 13:22:25 No.14118609 Report

Quoted By: >>14118878

OP should try to find a counter-example among 2 by 5 matrices. It should be easy to visualize this as a problem in finding 5 rays such that any vector can be generated as positive combinations of two different sets of the rays, but that if you choose one division of the rays into two sets, one of the two sets won't generate all vectors.

Anonymous

Anonymous Tue 18 Jan 2022 14:08:21 No.14118700 Report

Quoted By: >>14118878

>>14116758
Suppose $MX = b$

Then $M = (M_1,M_2) and X = (X_1,X_2) \rightarrow M_1X_1 = b_1, M_2X_2 = b_2 \rightarrow b_1+b_2 = b$
Or $(M_1X_1+M_2X_2) = b$

That is, if $MX = b$ , then every partition $(M_1,M_2)$ has a partition $(X_1,X_2)$
such that $M_1X_1 + M_2X_2 = b$ .

Introduce $X_3, X_4$ such that $M_1X_3 = b_2, M_2X_4 = b_1$
then $M_1(X_1+X_3) = b, M_2(X_2+X_4) = b$ .
$M_1X_3 = b_2 = M_2X_2 \rightarrow M_1X_3 = M_2X_2 \rightarrow X_3 = M_2*(M_1^{-1})X_2$ where
$(M_1^{-1})$ such that $M_1^{-1}M_1 = ID$ which is possible as long as
the columns of $M_1$ are linearly independent.

Similarly, $M_2X_4 = b_1 = M_1X_1 \rightarrow X_4 = M_1*(M_2^{-1})X_1$

Which shows that the solution to $M_1X = b, M_2X = b$ is in fact not dependent on b if there is a solution to
$MX = b$

Anonymous

Anonymous Tue 18 Jan 2022 15:26:40 No.14118878 Report

Quoted By:

>>14118700
Thanks ! But I didn't really understand what you were explaining.

>>14118609
Thanks, i did and it worked !

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