Linear algebra- Inverse of a linear mapping

  • Thread starter manuel325
  • Start date
  • #1
16
0

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 .
 

Answers and Replies

  • #3
16
0
Hint: ##I=-L^2-2L##.

hmm ,that doesn't tell me anything :confused:
 
  • #5
16
0
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:
 
  • #6
22,129
3,298
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

[tex]A\circ (B + C) = A\circ B + A\circ C[/tex]

and

[tex]A\circ (\alpha B) = \alpha (A\circ B)[/tex]

These properties are true, but it requires a separate proof.
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Those properties are, in fact, the definition of "linear transformation". That's all the proof you need.
 
  • #8
22,129
3,298
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.
 
  • #9
16
0
ok , I see .Thank you very much :smile:
 
  • #10
16
0
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
STEMucator
Homework Helper
2,075
140
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)##
 
  • #12
16
0
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
6,054
391
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
16
0
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
6,054
391
##-L - 2## is wrong to begin with. You cannot add (or subtract) numbers from linear maps. The correct expression is ##-L - 2I##.
 
  • #16
491
189
If there exist such an inverse map, what property does it have i.e. what can you tell about the multiplication of L and [itex]L^{-1}[/itex] ? Or the same question, what is the definition of the inverse map?
 
Last edited:
  • #17
16
0
##-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 .
 

Related Threads on Linear algebra- Inverse of a linear mapping

Replies
1
Views
730
  • Last Post
Replies
4
Views
981
Replies
1
Views
10K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
4
Views
8K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
4
Views
946
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
901
Top