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anonymous
 one year ago
if I have an mxn matrix A that does NOT have to be linearly independent and an affine subspace that's defined by (A^T)x = b, can it be proven that there always exists some combination of the columns of A which is part of that subspace?
anonymous
 one year ago
if I have an mxn matrix A that does NOT have to be linearly independent and an affine subspace that's defined by (A^T)x = b, can it be proven that there always exists some combination of the columns of A which is part of that subspace?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you guys help me with this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do you have so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i assume it is true, but i want to prove it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can think of a few examples where this statement is true, thus i assume it must be true for all examples, but this wouldn't be the mathematics category if I would leave it with this assumption.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0from your equation: \[\Large {A^T}x = b\] we can say that the column b is a linear combination of the column of A^T

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0namely, the column b is a linear combination of the rows of A

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0now we can say that the rank of A is equal to: \[\Large rank\left( A \right) \leqslant \min \left\{ {m,n} \right\}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is there something wrong with my browser or do you intend to use all these symbols? It is a pretty hard to read.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0for example if I have a 3 x 2 matrix, then all what I can say is that the rank of my matrix is less or equal 2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0in other words, exists some linearly independent rows of A, which correspond to the same number of ilinearly independent columns of A^T, such that they generate the I mage of A as linear map.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0and those rows of A are a basis of a linear subspace

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left[\begin{matrix}a11 & a12 & ... & a1n \\ a21 & ... & ... & ... \\ ... & ... & ... & ... \\ an1 & ... & ... & ann\end{matrix}\right] *\left(\begin{matrix}x1 \\ x2 \\ ... \\ xn\end{matrix}\right)\]
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