## 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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1. anonymous

can you guys help me with this?

2. anonymous

What do you have so far?

3. anonymous

nothing

4. anonymous

i assume it is true, but i want to prove it

5. Michele_Laino

what is A^T ?

6. anonymous

the transpose of A

7. Michele_Laino

ok!

8. anonymous

I 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.

9. Michele_Laino

from your equation: $\Large {A^T}x = b$ we can say that the column b is a linear combination of the column of A^T

10. Michele_Laino

columns*

11. Michele_Laino

namely, the column b is a linear combination of the rows of A

12. anonymous

yes

13. Michele_Laino

now we can say that the rank of A is equal to: $\Large rank\left( A \right) \leqslant \min \left\{ {m,n} \right\}$

14. anonymous

is there something wrong with my browser or do you intend to use all these symbols? It is a pretty hard to read.

15. Michele_Laino

for 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

16. anonymous

yes

17. Michele_Laino

in 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.

18. Michele_Laino

linearly*

19. Michele_Laino

and those rows of A are a basis of a linear subspace

20. anonymous

$\left[\begin{matrix}a11 & a12 & ... & a1n \\ a21 & ... & ... & ... \\ ... & ... & ... & ... \\ an1 & ... & ... & ann\end{matrix}\right] *\left(\begin{matrix}x1 \\ x2 \\ ... \\ xn\end{matrix}\right)$

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