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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
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.
We have to solve for the variables though right?
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/18-06-linear-algebra-spring-2010/video-lectures/lecture-7-solving-ax-0-pivot-variables-special-solutions/
basis for row space are the pivot rows basis for the col space are the pivot cols of the original matrix
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?
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......
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....
yes, that sounds right. I know the first vector in N(A) is [13/7 -1/2 -5/3 -2/7 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.
same idea for the col space, except you take the cols from A not rref(A)
got to go...
Oh noes :(.
wait crap.... :(
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.
oic you meant null basis :P