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KonradZuseBest ResponseYou've already chosen the best response.0
Basically I just need help finding how to set up the variables to be able to do this EQ @TuringTest @phi @satellite73 @farmdawgnation
 one year ago

phiBest ResponseYou've already chosen the best response.1
This question is asking about the column space, row space, null space you can find the null space by setting the bottom variable to 1 or 0
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Yes, I know, but I thought we need to find the variables first to find the null space and such... I was looking back on my past homeworks so I could be wrong, because I forgot how to do this mostly :P.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
We have to solve for the variables though right?
 one year ago

phiBest ResponseYou've already chosen the best response.1
I use a shortcut so often that the actual details are fuzzy. I just take the last 2 columns (the ones without a pivot) and negate all the entries in a "pivot row" then I put 1 and 0 in the remaining slots See http://ocw.mit.edu/courses/mathematics/1806linearalgebraspring2010/videolectures/lecture7solvingax0pivotvariablesspecialsolutions/
 one year ago

phiBest ResponseYou've already chosen the best response.1
basis for row space are the pivot rows basis for the col space are the pivot cols of the original matrix
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
From looking at my homework again I guess I don't really need to set up anything. It just seems that the last 2 columns have to do with it, which might be what you're getting it?
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Remark Observe that the basis vectors , , and in the last example are the vectors that result by successively setting one of the parameters in the general solution equal to 1 and the others equal to 0. hmm......
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
It also seems to me that they just take each row and make r1, r2, r3, etc and tha'ts the basis for the rows....
 one year ago

phiBest ResponseYou've already chosen the best response.1
yes, that sounds right. I know the first vector in N(A) is [13/7 1/2 5/3 2/7 1]
 one year ago

phiBest ResponseYou've already chosen the best response.1
It also seems to me that they just take each row and make r1, r2, r3, etc and tha'ts the basis for the rows.... the rows with a pivot.
 one year ago

phiBest ResponseYou've already chosen the best response.1
same idea for the col space, except you take the cols from A not rref(A)
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I know the first vector in N(A) is [13/7 1/2 5/3 2/7 1] isn't a row? :( a col maybe that's negated :P.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
oic you meant null basis :P
 one year ago
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