Vector spaces homework question (rowspace and nullspace)

murielg
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Homework Statement



Write x=(6,-1,-2)T as x=y+z where y belongs to null A and z belongs to row A

A=[1,3,1;2,6,2;-2,-5,0;1,4,3]

Homework Equations


The main question asks to find all the fundamental subspaces and their dimensions, which I have already found, and then asks me to find the question i posted above.

the basis for the row space is
{[1,3,1]T, [-1,-5,0]T}

the basis for the nullspace is
(5,-2,1)T

The Attempt at a Solution



I don't really know what the question is asking me to do... or how to begin to approach this problem.
If someone could please give me a hint on what they want me to do.

Thanks a lot
 
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Can you write (6,-1,-2) in terms of (1,3,1), (-1,-5,0) and (5,-2,1)?
 
is this like a change of basis?
im confused :S and I am sure it's really simple but I really want to understand what this means
im going to keep giving this some thought, thanks for the fast reply
 
Find a, b, and c so that (6,-1,-2)= a(1,3,1)+ b(-1,-5,0)+ c(5,-2,1). Once you have done that, y= a(1, 3, 1)+ b(-1, -5, 0) and z= c(5, -2, 1).
 
I found X1=1, x2=-3, x3=-2

and then (6,-1,-2)T= (2,-11,-2)T + (4,10,0)T
as x = y + z where Y belongs to the nullspace and z to the rowspace

Thanks guys!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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