November 23rd, 2008, 02:39 PM  #1 
Newbie Joined: Nov 2008 Posts: 4 Thanks: 0  Homework question
The directions state: Find the solution space of the following systems of linear homogeneous equations: xy+zw=0 2x+yz+2w=0 2y+3z+w=0 Is this the same as finding the solutions, because I did that and got (z, 3z, 3z, z). But I don't know what a solution space is. Help. Thanks you so much. 
November 24th, 2008, 03:33 PM  #2 
Newbie Joined: Nov 2008 Posts: 4 Thanks: 0  Re: Homework question
To answer my own question: the solution space is actually the subspace you get. For instance, this problem yields a one dimensional subspace. 
December 2nd, 2008, 06:45 PM  #3 
Newbie Joined: Nov 2008 Posts: 4 Thanks: 0  Re: Homework question
I don't think that solution set is right. The vector 1,3,3,1 does not give you 0 when used as weights in the original equation. It is a system of 4 unknowns in 3 rows there has to be a free variable somewhere. The solution (1,3,3,1) does not satisfy ax=0. That is if you still care about this problem :P, unless im missing something.. 
December 7th, 2008, 07:04 AM  #4 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: Homework question
Oh, wow, I completely missed this problem... Anyway, yes, the vector (1,3,3,1) forms a basis for the vector space. It isn't the solution itself z is the free variable. So, undermine, you are missing something 
April 15th, 2015, 08:35 PM  #5 
Newbie Joined: Apr 2015 From: usa Posts: 1 Thanks: 0 
Thanks! Looking forward to it. I think I've spend more time watching your show this at work than actually working! Fantastic stuff!

April 16th, 2015, 04:17 PM  #6  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,170 Thanks: 869  Quote:
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April 16th, 2015, 05:04 PM  #7  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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