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April 23rd, 2013, 08:30 PM  #1 
Newbie Joined: Apr 2013 Posts: 11 Thanks: 0  Finding basis for an operator of particular form http://imageshack.us/photo/myimages/694/piczz.jpg/ Sorry I am unable to write the problem in the text box. It has some symbols. Please find the attached link. Can anyone help me out? Thanks a lot in advance. 
April 24th, 2013, 07:07 PM  #2 
Newbie Joined: Apr 2013 Posts: 11 Thanks: 0  Re: Finding basis for an operator of particular form
No help yet?

April 25th, 2013, 07:08 AM  #3 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Finding basis for an operator of particular form
A lot of people will not open files posted by people they do not know. There certainly isn't any reason why you could not have typed the problem in even if you don't use latex. Let T be a nonzero operator on C^2 such that T^2= 0. Show that there exist a basis for C^2 in which the matrix for T is [ 0 1] [0 0] Since T is nonzero there exist at least one vector, v, such that Tv= u is not 0. Since T^2= 0 , Tu= T^2v= 0. Construct the matrix representing T using those two vectors as a basis for C^2. 
April 25th, 2013, 10:54 AM  #4 
Newbie Joined: Apr 2013 Posts: 11 Thanks: 0  Re: Finding basis for an operator of particular form
Thanks HallsofIvy, thanks a lot. I sould have realized I could type that way. According to your explanation. Tv=u where u is not 0 and Tu=0. Looking at the matrix [0 1] we can say Tu=(0,0) and Tv=(1,0) [0 0] Thus u=(1,0). Since, Tv=u, we are left with finding v which makes Tv=(1,0) May I ask what would be v in case if I am correct. Thanks a lot again HallsofIvy. 
April 25th, 2013, 12:15 PM  #5 
Newbie Joined: Apr 2013 Posts: 11 Thanks: 0  Re: Finding basis for an operator of particular form
Sorry, HallsofIvy I guess I was wrong in the concept. Disregard the earlier post please. I have Tu=0 Tv=u Let (u,v) be the basis of C^2. In that case the matrix would have the given form since Tu=0.u+0.v and Tv=1.u+0.v So the nonzero basis would be (u,v) s.t Tu=0 and Tv=u. Can you define any such T? Thanks again. 

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