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 February 27th, 2011, 12:47 PM #1 Newbie   Joined: Feb 2011 Posts: 10 Thanks: 0 Kernel of a linear map over a vector space k Hi Let $f$ be a linear map over a vector space $k$ where $k$ is generated by $x_o,...., x_d$ and $f^n=0$ where $n>0$ Also,$x_ix_ j + x_ jx_i=0$ for each $i , j.$ Show that $\ker (f)$ is non-zero. Please give me just a hint.
February 27th, 2011, 01:50 PM   #2
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Re: Kernel of a linear map over a vector space k

Quote:
 Originally Posted by Saba Also,$x_ix_ j + x_ jx_i=0$ for each $i , j.$
Would you please clarify what you mean by this ? There is no multiplication on a standard vector space, only an external law (Field, Vector space) -> Vector space. Are you working with an algebra or something ?

 February 27th, 2011, 02:00 PM #3 Newbie   Joined: Feb 2011 Posts: 10 Thanks: 0 Re: Kernel of a linear map over a vector space k $f$is a linear map on a $k$-vector space M with $f^n=0$ for some $n>0$ $k$ is a field $k^d$is $k$-algebra generated by $x_1,...x_d$ satisfy the relation $x_ix_j+x_jx_i=0 , i,j=1,...,d$
 February 27th, 2011, 02:01 PM #4 Newbie   Joined: Feb 2011 Posts: 10 Thanks: 0 Re: Kernel of a linear map over a vector space k Thanx in advance
 February 27th, 2011, 03:02 PM #5 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 Re: Kernel of a linear map over a vector space k Hi Saba, your notaion is confusing. Infact you keep changing it. The good thing is that you don't need anything of the algebra stuff for your result: Any nilpotent endomorphism f (i.e. $f^n=0$) of a nonzero vector space (or basically any structure you can think of) has this property. As a hint: Show the stronger result: The composition of any two injective morphisms is injective. Generalise to n morphisms and conclude your claim.
 February 27th, 2011, 03:08 PM #6 Newbie   Joined: Feb 2011 Posts: 10 Thanks: 0 Re: Kernel of a linear map over a vector space k if the map is injective , then ker(f) =0 But here ,show that ker f is NON-zero
February 28th, 2011, 03:27 AM   #7
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Re: Kernel of a linear map over a vector space k

Quote:
 Originally Posted by Saba But here ,show that ker f is NON-zero
Well, you just wanted a hint. I would suggest to assume for contradiction that f is injective, then show that$f^n$ is injective. But this is absurd, as $f^n=0$ and your vector space is non-zero.

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