Question on Null and Image Space matrices wee

mr_coffee
Messages
1,613
Reaction score
1
Hello everyone! I'm alittle confused, my professor found the Null space after some students didn't get how to do it, such as myself. Here is the image and problem:
http://img497.imageshack.us/img497/4161/lastscan9es.jpg
Thats the row reduced forum of the orginal matrix. The orginal matrix was:
-1 4 1
7 -9 0
10 3 -3
-9 1 2

he added a column of 0's so it wuold make it a square matrix, then he row reduced to get the matrix in the link.
I understand where the Null space of A is:
9/37
7/37
1
0

that makes sense to me, but then he adds in:
d =
0
0
0
1
why is that also a set of vectors in the Null space? d is just all 0's, how can he have a 1 also?


Then I'm lost on what I'm suppose to do to find the image space, you can see my work in the image above. What do i do after i set them equal to r,s,t,u? Thanks! :smile:
 
Last edited by a moderator:
Physics news on Phys.org
I was going to make a new thread, but its the same question different problem. Did I find the correct image space on this problem?
http://img500.imageshack.us/img500/2195/lastscan7tn.jpg
THanks!
 
Last edited by a moderator:
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...
Back
Top