Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $L: V\to V$ be an operator in a finite-dimensional vector space $V$ over $R$. For any $n \geq 0$, let $K_n = \ker (L^n)$, $I_n = \mathrm{Im}(L^n)$.

(a) Prove that there exists $N$ such that for all $n \geq N$, we have $K_n=K_N$ , $I_n= I_N$.

(b) Denote $K = K_N$, $I = I_N$, where $N$ is the same as above. Prove that $LK$ is contained in $K$, and $LI$ is contained in $I$, and the restriction of $L$ to $K$ is nilpotent, restriction of $L$ to $I$ is invertible.

(c) Prove that $V = K \oplus I$.

(For part c, We assume without proof that if $p \in R[x]$ is the characteristic polynomial of $L$, then $p(L) = 0$ , how to proceed?)

share|improve this question
2  
Just view matrices as $m$-tuples of column vectors in $L$. –  Hagen von Eitzen Dec 13 '12 at 22:40
    
??????????????????????????????????????? –  Mike Dec 14 '12 at 3:47
    
What have you done? Where are you stuck? –  Olivier Bégassat Dec 14 '12 at 4:07
    
Please don't completely change your question. Ask a new one instead. –  espen180 Dec 14 '12 at 13:35

1 Answer 1

Try thinking about how the $R$-action acts on $L$ and $R$.

Look at the $R$-homomorphism sending $(l_1,...,l_n)\in L^n$ to $(l_1|...|l_n)=A\in R$, such that $l_i$ is the $i$-th column vector of $A$.

Do you see how this map gives an isomorphism of $R$-modules?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.