anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
can you guys help me with this?
anonymous
  • anonymous
What do you have so far?
anonymous
  • anonymous
nothing

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More answers

anonymous
  • anonymous
i assume it is true, but i want to prove it
Michele_Laino
  • Michele_Laino
what is A^T ?
anonymous
  • anonymous
the transpose of A
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • 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.
Michele_Laino
  • 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
Michele_Laino
  • Michele_Laino
columns*
Michele_Laino
  • Michele_Laino
namely, the column b is a linear combination of the rows of A
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
now we can say that the rank of A is equal to: \[\Large rank\left( A \right) \leqslant \min \left\{ {m,n} \right\}\]
anonymous
  • anonymous
is there something wrong with my browser or do you intend to use all these symbols? It is a pretty hard to read.
Michele_Laino
  • 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
anonymous
  • anonymous
yes
Michele_Laino
  • 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.
Michele_Laino
  • Michele_Laino
linearly*
Michele_Laino
  • Michele_Laino
and those rows of A are a basis of a linear subspace
anonymous
  • 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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