Do A and Uinv(A)U Share Eigenvalues?

[Ed Quanta]

A is an operator and Uinv is the inverse of operator U

How am I to show that A and Uinv(A)U have the same eigenvalues? Must U be unitary for this to be true?

And if the eigenvectors of A are (Psi n), what are the eigenvectors of Uinv(A)U?

Help anyone? Not sure what to do here.

 

[Wong]

Unitarity of U is not required. To prove your assertion, try to start with the equation Av=\lambda v. How may you manipulate this equation to obtain U^{-1}AU?

 

[Ed Quanta]

So can I just do something like following:?

We know Av=\lambda 2 v

We then let Uv=\lambda 1 v

So then Av \lambda1=\lambda 2 \lambda1 v

Then since we know
U^{-1}U=1, we know

U^{-1}v=v/(\lambda 1 )

ThusU^{-1}AUv=\lambda 2 v

 

[Dr Transport]

Not quite, but close. Multiply the original equation on the left by U then insert 1 = U^{-1}U and work from there...

 

[Ed Quanta]

Can you explain to me why what I did is wrong? And when you say to multiply thee original equation by U, and then use the fact that 1=U multiplied by its inverse, it seems that the order of terms in the equation is altered from the expected result U^{-1}AU

And I know when it comes to matrix multiplication AB does not equal BA, so I am a tad confused as to what you want me to do. I apologize for my slowness.

 

[Wong]

To get you started,

Av=\lambda v
U^{-1}Av=\lambda U^{-1}v

Now how may you manipulate the last equation to get U^{-1}AU?