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

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

Thus$$U^{-1}AUv=\lambda 2 v$$

 

[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 $$U^{-1}AU$$?