# Linear algebra- Inverse of a linear mapping

1. Jul 6, 2013

### manuel325

1. The problem statement, all variables and given/known data
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 .

2. Jul 6, 2013

### micromass

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

3. Jul 6, 2013

### manuel325

hmm ,that doesn't tell me anything

4. Jul 6, 2013

### micromass

Staff Emeritus
Factor out $L$ in the right hand side.

5. Jul 6, 2013

### manuel325

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

6. Jul 6, 2013

### micromass

Staff Emeritus
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.

7. Jul 6, 2013

### HallsofIvy

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

8. Jul 6, 2013

### micromass

Staff Emeritus
I wouldn't say it's the definition. It still requires a proof.

9. Jul 6, 2013

### manuel325

ok , I see .Thank you very much

10. Jul 7, 2013

### 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.

11. Jul 7, 2013

### Zondrina

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 :

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

12. Jul 7, 2013

### manuel325

Ok so $L^{-1}I = L^{-1}L(-L-2)$ yields $I \circ (-L-2)=-L-2$ right ?? . Thanks

13. Jul 7, 2013

### voko

No, you cannot do that. You are proving here that $L^{-1}$ exists, so you cannot assume it exists in the proof!

14. Jul 7, 2013

### manuel325

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.

15. Jul 7, 2013

### voko

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

16. Jul 7, 2013

### 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?

Last edited: Jul 7, 2013
17. Jul 7, 2013

### manuel325

It makes sense now ,thanks .