How to show that linear map is invertible?

Question

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.

score 7 · Accepted Answer · answered Jan 29 '13 at 21:42

7

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. $v=0$

answered Jan 29 '13 at 21:42

Emanuele Paolini

21,447

score 4 · Answer 2 · answered Jan 29 '13 at 21:44

4

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

answered Jan 29 '13 at 21:44

M.H

11,498
3
30
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score 2 · Answer 3 · answered Jan 29 '13 at 21:43

2

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.

answered Jan 29 '13 at 21:43

Seirios

33,157

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

Thomas · Answer 4 · 2013-01-29T21:49:42.253

2

For the first one you have $$ L(L + 2I) = -I. $$ For the second one you have $$ (I-L)(I+L)(I+L^2) = \dots = I. $$

edited Jan 29 '13 at 21:49

answered Jan 29 '13 at 21:44

Thomas

43,555

Andy Brandi · Answer 5 · 2013-01-29T23:01:23.843

2

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 $L$.

edited Jan 29 '13 at 23:01

answered Jan 29 '13 at 22:00

Andy Brandi

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I very much like this approach, it is very natural. – Ragib Zaman Jan 30 '13 at 01:48

score 2 · Answer 6 · edited Jan 29 '13 at 23:08

2

For the second you should use the property: $(1-x)(1+x+x^2)=1-x^3$.

edited Jan 29 '13 at 23:08

Clive Newstead

64,925

answered Jan 29 '13 at 22:39

a k

29

score 2 · Answer 7 · answered Jan 30 '13 at 01:48

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 $n=3.$

How to show that linear map is invertible?

7 Answers7