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$f$ 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 $x_1,\ldots,x_d$ satisfy the relation $x_i x_j + x_j x_i =0$ for all $i,j=1, \ldots, d$.

Show that $\ker(f)$ is non-zero.

Thank you!

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What have you tried so far? –  Aryabhata Feb 27 '11 at 23:13
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what does k^d have to do with anything? –  Eric O. Korman Feb 27 '11 at 23:13
    
let x in k^d, then x = linear combination of x_i, then x-(sum r_ix_i) =0 then f (x-sum)= 0 ... but it does not work.. I do not know how can i use f^n =0 and the condition –  saba Feb 27 '11 at 23:20
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1 Answer

Hint: Choose $n$ in the question such that $f^{n-1}$ is not zero, and notice that $f(f^{n-1}(M))=\{0\}$.

Alternatively, prove that in general a composition of injective maps is injective.

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Please , could you explain more? does that mean ker(f)=f^(n-1)[M]?? –  saba Feb 27 '11 at 23:40
    
@saba: x is in ker(f) if and only if f(x)=0. If A is a subset of M such that f(A)={0}, meaning that f(x)=0 for all x in A, then every element of A is in ker(f); that is, A is a subset of the kernel. $$ $$ As for the second suggestion, have you seen that a linear map has kernel {0} if and only if it is injective (one-to-one)? If so, it suffices to show that f is not injective. You can do this by contraposition. If f is an injective map, then so are $f\circ f$, $f\circ f\circ f$, $f\circ f\circ f\circ f$, etc. The zero map is not injective. –  Jonas Meyer Feb 27 '11 at 23:58
    
THANK YOU SOO MUCH –  saba Feb 28 '11 at 0:12
    
@saba: If you like the answer (including the additional comment-it would appear so) please accept it by clicking the check mark next to this answer. It will prevent it from popping up in the future. –  Ross Millikan Mar 30 '11 at 3:22
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