Linear algebra- Inverse of a linear mapping

AI Thread Summary
The discussion revolves around proving the invertibility of a linear mapping L given the equation L^2 + 2L + I = 0. Participants clarify that L can be manipulated like a number in this context, emphasizing the properties of linear transformations. The correct expression for the inverse is identified as L^(-1) = -L - 2I, correcting misconceptions about adding scalars to linear maps. The conversation highlights the importance of understanding the definitions and properties of linear mappings to solve the problem effectively. Overall, the participants work through the confusion surrounding the operations involving linear mappings and scalars.
manuel325
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Homework Statement


Let L: V →V be a linear mapping such that L^2+2L+I=0, show that L is invertible (I is the identity mapping)
I have no idea how to solve this problem or how to start,I mean this problem is different from the ones I solved before, the answer is "The inverse of L is -L-2 "
If someone please can explain to me how to solve this ,would be great :cool:. Thanks in advance .
 
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Hint: ##I=-L^2-2L##.
 
micromass said:
Hint: ##I=-L^2-2L##.

hmm ,that doesn't tell me anything :confused:
 
manuel325 said:
hmm ,that doesn't tell me anything :confused:

Factor out ##L## in the right hand side.
 
micromass said:
Factor out ##L## in the right hand side.
hmm ok you mean I=L(-L-2) but can you operate with L^2 like it was a number?? , isn't L^2 =L°L ?? I'm confused:confused:
 
Very good remark!

The answer you can operate with it like it was a number, but that's perhaps not obvious.
The property I'm using here is that

A\circ (B + C) = A\circ B + A\circ C

and

A\circ (\alpha B) = \alpha (A\circ B)

These properties are true, but it requires a separate proof.
 
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Those properties are, in fact, the definition of "linear transformation". That's all the proof you need.
 
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HallsofIvy said:
Those properties are, in fact, the definition of "linear transformation". That's all the proof you need.

I wouldn't say it's the definition. It still requires a proof.
 
ok , I see .Thank you very much :smile:
 
  • #10
Hmm I guess there's something that's still not clear for me , the first property of composition you wrote works for three linear mappings right?, but in this case we have two linear mappings and a number :L°(-L-2) I know I'm wrong somewhere but I don't know where ,I'm confused :confused: .Any help would be appreciated.
 
  • #11
micromass said:
Hint: ##I=-L^2-2L##.

Micro stated this here, you should be able to see that :

##I = L(-L-2)##

Then multiplying both sides by ##L^{-1}## on the left yields :

##L^{-1}I = L^{-1}L(-L-2)##
##L^{-1} = -(L+2)##
 
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  • #12
Zondrina said:
Micro stated this here, you should be able to see that :

##I = L(-L-2)##

Then multiplying both sides by ##L^{-1}## on the left yields :

##L^{-1}I = L^{-1}L(-L-2)##
##L^{-1} = -(L+2)##
Ok so ##L^{-1}I = L^{-1}L(-L-2)## yields ##I \circ (-L-2)=-L-2## right ?? . Thanks
 
  • #13
Zondrina said:
Micro stated this here, you should be able to see that :

##I = L(-L-2)##

Then multiplying both sides by ##L^{-1}## on the left yields :

No, you cannot do that. You are proving here that ##L^{-1}## exists, so you cannot assume it exists in the proof!
 
  • #14
voko said:
No, you cannot do that. You are proving here that ##L^{-1}## exists, so you cannot assume it exists in the proof!

You are right !:smile: could you explain please why ## L\circ(-L-2)=-L^{2}-2L## ?? if "2" was a linear mapping then the properties would work for this, right? but 2 is a scalar.
 
  • #15
##-L - 2## is wrong to begin with. You cannot add (or subtract) numbers from linear maps. The correct expression is ##-L - 2I##.
 
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  • #16
If there exist such an inverse map, what property does it have i.e. what can you tell about the multiplication of L and L^{-1} ? Or the same question, what is the definition of the inverse map?
 
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  • #17
voko said:
##-L - 2## is wrong to begin with. You cannot add (or subtract) numbers from linear maps. The correct expression is ##-L - 2I##.

It makes sense now ,thanks .
 
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