# Eigenvalues question.

1. Jun 12, 2009

### Batman2

1. The problem statement, all variables and given/known data
Let B be an invertible matrix

a.) Verify that B cannot have zero as an eigenvalue.

b.) Verify that if $$\lambda$$ is an eigenvalue of B, then $$\lambda$$$$^{-1}^$$ is an eigenvalue of B$$^{-1}$$.

2. Relevant equations
Bv = $$\lambda$$v, where v$$\neq$$0

3. The attempt at a solution
a.) I'm pretty sure that I need to manipulate the eigenvalues definition above so that I end up with v = 0, thus contradicting the definition.

What I have so far is:
Bv = $$\lambda$$v
B$$^{-1}$$Bv = B$$^{-1}$$$$\lambda$$v
Iv = B$$^{-1}$$$$\lambda$$v
v = $$\lambda$$B$$^{-1}$$v
If $$\lambda$$ = 0
v = 0, which contradicts the definition for eigenvalues where v$$\neq$$0
Therefore $$\lambda$$$$\neq$$0

I think I am missing some crucial steps and kinda jumped ahead in my working. Am I on the right track? How can I approach this problem properly?

b.) I'm thinking of a similar approach for b.), where I would need to use the above definition and multiply through by the inverse B, and then maybe take the reciprocal of lambda.

However I'm not sure given that I can't do the first part yet. Any help would be much appreciated.

Daniel

2. Jun 12, 2009

### rock.freak667

Normally, I'd say that looks like it'd work, but I am not sure as I only did this topic like 3 years ago.

and for the first one, I'd just use what the product of the eigenvalues would give in relation to B and then show how it would contradict the statement of what B is given $\lambda=0$

3. Jun 12, 2009

### Hurkyl

Staff Emeritus
I suspect that it's that eerie feeling you get when you really start to understand something, and are surprised at how straightforward it all seems!

Just to organize what you've done a little bit (and probably with too much detail), your arithmetic has shown

Given a vector v, scalar $\lambda$, and matrix B:

If $Bv = \lambda v$ and B is invertible
Then $v = \lambda B^{-1} v$

If, in addition, $\lambda = 0$,
Then $v = 0$
And you have correctly argued
If B is invertible, $\lambda$ is an eigenvalue, and $\lambda = 0$
And correctly concluded
If B is invertible, $\lambda$ is an eigenvalue
Then $\lambda \neq 0$​

4. Jun 13, 2009

### HallsofIvy

Staff Emeritus
For (b), if $Bv= \lamba v$ then, as you have, $B^{-1}Bv= B^{-1}\lambda v$ or $v= \lambda B^{-1}v$. Now divide on both sides by $\lambda$.

5. Jun 13, 2009

### MLeszega

I remember doing these problems. I think this is how I did part a:

Bv=$$\lambda$$v
Bv-$$\lambda$$v=0
(B-$$\lambda$$)v=0

If $$\lambda$$=0 then Bv=0 which can only happen if B is not invertible, which is a contradiction.

As for part b: If Bv= $$\lambda$$v then B-1v=$$\lambda$$-1v
Then B-1v-$$\lambda$$-1v=0
(B-1-$$\lambda$$-1)v=0 and you now have the definition of an eigenvalue, where $$\lambda$$-1 is an eigenvalue for B-1.

Last edited: Jun 13, 2009
6. Jun 13, 2009

### Hurkyl

Staff Emeritus
You know, you guys shouldn't be doing his homework for him. :grumpy: