Can an Invertible Matrix Have Zero as an Eigenvalue?

[Batman2]

Homework Statement

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$$^{-1}$$.

Homework Equations

Bv = $$\lambda$$v, where v$$\neq$$0

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

 

[rock.freak667]

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.

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 $\lambda=0$

 

[Hurkyl]

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?

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 $\lambda = 0$
Then we have a contradiction​

And correctly concluded

If B is invertible, $\lambda$ is an eigenvalue
Then $\lambda \neq 0$​

 

[HallsofIvy]

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 $\lambda$.

 

[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.

 

[Hurkyl]

You know, you guys shouldn't be doing his homework for him.

Fortunately, from the opening poster's comments, I'm sure he had worked it out already.