Eigen value proof

1. Dec 10, 2012

pyroknife

Let A be an invertible matrix. Show that if λ is
an eigenvalue of A, then 1/λ is an eigenvalue of
A^−1.

det((A-λI))
det((A-λI)^-1)
=det(A^-1 - λ^-1 * I)
=det(A-1-1/λ*I)

Is this enough to show that?

Another question I have is:
Let A be an n × n matrix. Show that A is not
invertible if and only if λ = 0 is an eigenvalue
of A.

Not sure how to approach this prob.
det(A-λI)=0
det(A-0I)=0
det(A)=0

But idk how to show what the problem is asking.

2. Dec 10, 2012

Staff: Mentor

(Above) No... Is this some kind of new distributive property for exponents? Kind of like (a+ b)2 = a2 + b2?
No. Start with what it means for λ to be an eigenvalue of A.
And what it means for λ-1 to be an eigenvalue of A-1.

Last edited: Dec 10, 2012
3. Dec 10, 2012

Staff: Mentor

This is an "if and only if" proof, so you have to prove two propositions:
1. If λ = 0 is an eigenvalue of A, then A is not invertible.
2. If A is not invertible, then λ = 0 is an eigenvalue of A.

For the first one, start with what it means for a number to be an eigenvalue of a matrix.
For the second one, what can you say about a matrix that does not have an inverse?

4. Dec 10, 2012

pyroknife

Let A be an invertible matrix. Show that if λ is
an eigenvalue of A, then 1/λ is an eigenvalue of
A^−1.
det((A-λI))
det((A-λI)^-1)
=det(A^-1 - λ^-1 * I)
=det(A-1-1/λ*I)

if λ is an eigenvalue of A then the following is true:
Av=λv where v=eigenvec
A^-1*Av=A^-1*(λv)=A^-1*Av=Iv
A^-1*Av=Iv=v
A^-1*λv=Iv=v
A^-1*v=(1/λ)*v

5. Dec 10, 2012

Staff: Mentor

You can get rid of this stuff - it's wrong.
And v has to be nonzero.
It seems like you're going around in a circle here.
Just from the left side above, we have A-1Av = Iv = v
From where you started, you want to end up with this statement:

A-1u = (1/λ)u

6. Dec 10, 2012

pyroknife

1. If λ = 0 then det(A-λI)=0=det(A-0)=det(A) so A is not invertible
2. det(A)=0 if it doesn't have an inverse. so that means λ = 0