# Linear algebra- Inverse of a linear mapping

## 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 . Thanks in advance .

Hint: ##I=-L^2-2L##.

Hint: ##I=-L^2-2L##.

hmm ,that doesn't tell me anything

hmm ,that doesn't tell me anything

Factor out ##L## in the right hand side.

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

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.

1 person
HallsofIvy
Homework Helper
Those properties are, in fact, the definition of "linear transformation". That's all the proof you need.

1 person
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

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 .Any help would be appreciated.

STEMucator
Homework Helper
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)##

1 person
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

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!

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 ! 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.

##-L - 2## is wrong to begin with. You cannot add (or subtract) numbers from linear maps. The correct expression is ##-L - 2I##.

1 person
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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##-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 .