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a) Let $L:V \to V$ be a linear map such that $L^2 + 2L + I = 0$, show that $L$ is invertible.
b) Let $L:V \to V$ be a linear map such that $L^3 = 0$, show that $I - L$ is invertible.

Here, $I$ is identity mapping.

For first part, I know that $L^2 + 2L + I = (L+I)^2 = 0$, if $v\in V$ then $(L+I)^2 v = (L+I)(L(v) + v)) = 0$ so $L(v) + v$ is in null space of $L+I$, from here how do I show that $0$ is only in null space of $L$.

I don't need exact solution. Hints would suffice.

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7 Answers

up vote 7 down vote accepted

a) Suppose that $Lv=0$ then $0 = L^2v +2Lv+v = L0 + 0 + v = v$ i.e. $v=0$.

b) Suppose $(I-L)v=0$ i.e $Lv = v$ hence $0 = L^3 v =v$ i.e. $v=0$

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$L^2 + 2L + I = O$ then $-L^2-2l=I$ then $L(-L-2)=I$ therfore $$L^{-1} =-L-2$$

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Hint: You can show that if $L$ is an endomorphism and $P$ is a polynomial such that $P(0) \neq 0$ and $P(L)=0$, then $L$ is invertible.

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i guess this is little complicated, since I haven't gone far as to read endomorphism. Still thank you very much for your effort. –  hasExams Jan 29 '13 at 21:54
    
@testuser: In fact, it is quite simple. For your example $L^2+2L+I=0$, just notice that $-L(L+2I)=I$. –  Seirios Jan 29 '13 at 21:57
    
Yep, it is :) ... only I could have realized sooner!! –  hasExams Jan 29 '13 at 22:03
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For the first one you have $$ L(L + 2I) = -I. $$ For the second one you have $$ (I-L)(I+L)(I+L^2) = \dots = I. $$

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Yet another approach for a), although it might be overkill:

Let $p(X) := X^2 + 2X + 1$. As $p(L)=0$, the minimal polynomial of $L$ divides $p$. Conclude that $0$ is not an eigenvalue of $L$.

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I very much like this approach, it is very natural. –  Ragib Zaman Jan 30 '13 at 1:48
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For the second you should use the property: $(1-x)(1+x+x^2)=1-x^3$.

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The same proposition deals with both parts:

If $L^n=0$ for some $n\in \mathbb{N}$ then $I-L$ is invertible. To prove this, check that $I+L+L^2 + \cdots + L^{n-1}$ is the inverse.

a) $(L+I)^2=0$ so $I-(L+I) = -L$ is invertible, and hence so is $L.$

b) is precisely the case $n=3.$

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