Eigenvalues and Eigenvectors of Invertible Linear Operators and Matrices

[hkus10]

Let L : V>>>V be an invertible linear operator and let lambda be an eigenvalue of L with associated eigenvector x.
a) Show that 1/lambda is an eigenvalue of L^-1 with associated eigenvector x.
For this question, the things I know are that L is onto and one to one. Therefore, how to prove this question?

b) state the analogous statement for matrices. What does "state" the analogous statement mean?

Thanks

 

[Mark44]

Let L : V>>>V be an invertible linear operator and let lambda be an eigenvalue of L with associated eigenvector x.
a) Show that 1/lambda is an eigenvalue of L^-1 with associated eigenvector x.
For this question, the things I know are that L is onto and one to one. Therefore, how to prove this question?

What does it mean to say that λ is an eigenvalue, and x an eigenvector of L?

b) state the analogous statement for matrices. What does "state" the analogous statement mean?

L(x) corresponds to Ax, where A is a matrix representation of the operator L. L is invertible, so what can you say about A?

 

[hkus10]

What does it mean to say that λ is an eigenvalue, and x an eigenvector of L?

a) This is what I get for part(a)let n=lambda.
Since r is an eigenvalue of L, Lx=nx.
Since the transformation is invertible, (L^-1)Lx=(L^-1)nx.
==> Ix=r(L^-1)x, where I=indentity matrix
At this point, I want to divide both sides by r. However, how can I be sure r is not equal to zero?

Thanks

 

[Dick]

a) This is what I get for part(a)let n=lambda.
Since r is an eigenvalue of L, Lx=nx.
Since the transformation is invertible, (L^-1)Lx=(L^-1)nx.
==> Ix=r(L^-1)x, where I=indentity matrix
At this point, I want to divide both sides by r. However, how can I be sure r is not equal to zero?

Thanks

You seem to be using both n and r for lambda. But if the eigenvalue were 0, would L be invertible?

 

[hkus10]

You seem to be using both n and r for lambda. But if the eigenvalue were 0, would L be invertible?

Yes, this is a typo. It seems to me that if the eigenvalue is 0, L is not invertible. However, I cannot conceptualize it.

Could you please explain it ?

Thanks

 

[Dick]

Yes, this is a typo. It seems to me that if the eigenvalue is 0, L is not invertible. However, I cannot conceptualize it.

Could you please explain it ?

Thanks

If the eigenvalue is 0, then L can't be invertible. If x is an eigenvector (hence not the zero vector) and has eigenvalue 0, then L(x)=0*x=0. But L^(-1)(0) must be 0, it can't be x since L^(-1) is linear. So if L is invertible, you can assume all eigenvalues are not zero.