Linear algebra- Inverse of a linear mapping

[manuel325]

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 .

 

[micromass]

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

 

[manuel325]

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

hmm ,that doesn't tell me anything

 

[micromass]

hmm ,that doesn't tell me anything

Factor out $L$ in the right hand side.

 

[manuel325]

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

 

[micromass]

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.

 

[HallsofIvy]

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

 

[micromass]

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.

 

[manuel325]

ok , I see .Thank you very much

 

[manuel325]

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]

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 :

$L^{-1}I = L^{-1}L(-L-2)$
$L^{-1} = -(L+2)$

 

[manuel325]

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

 

[voko]

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!

 

[manuel325]

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.

 

[voko]

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

 

[JorisL]

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?

 

[manuel325]

$-L - 2$ 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 .