One more Quantum Matrix question

[Ed Quanta]

Let A be a Hermitian nxn matrix. Let the column vectors of the nxn matrix S be comprised of the orthnormalized eigenvectors of A

Again, Sinv is the inverse of S

a) Show that S is unitary
b) Show that Sinv(A)S is a diagonal matrix comrpised of the eigenvalues of A

No idea how to start this one off.

 

[Wong]

a) U is a unitary matrix <=> U*U = I, where "*" denotes conjugate transpose <=> $$\sum_{j} u_{ji}^{*}u_{jk} = \delta_{ik}$$ <=> $$u_{i}^{*}u_{k}=\delta_{ik}$$, where $$u_{i}$$ is the ith column of U. The last relation implies orthogonality of columns of U.

b)This one needs a little thought. If u is an eigenvector of A, then $$Au=\lambda u$$. Then what is AS? Remember that each column of S is just an eigenvector of A. Also note that Sinv*S=I.

 

[Ed Quanta]

a) U is a unitary matrix <=> U*U = I, where "*" denotes conjugate transpose <=> $$\sum_{j} u_{ji}^{*}u_{jk} = \delta_{ik}$$ <=> $$u_{i}^{*}u_{k}=\delta_{ik}$$, where $$u_{i}$$ is the ith column of U. The last relation implies orthogonality of columns of U.

b)This one needs a little thought. If u is an eigenvector of A, then $$Au=\lambda u$$. Then what is AS? Remember that each column of S is just an eigenvector of A. Also note that Sinv*S=I.

Sorry, I am still not sure how to find AS without knowing the eigenvectors of A.

 

[Wong]

Sorry, I am still not sure how to find AS without knowing the eigenvectors of A.

First try to think about what you want to prove. That is, $$S^{-1}AS=D$$, where D is a diagonal matrix. This is equivalent to proving AS=DS, where D is diagonal. Now each column of S is an eigenvector of A. So A acting on S should produce something quite simple. (Try to think of what is the defining eigenvalue equation for A.) May you put the result in the form DS, where D is a diagonal matrix?

 

[erraticimpulse]

Wong Wrong

I have doubts that either of you guys will read this anytime soon. I had this same problem and the conclusion that Wong tried to provide is incorrect. Instead of $$AS=DS$$ it's actually $$AS=SD$$. The product DS will produce the correct entries along the diagonal but false elsewhere (really think about what you're doing here). But if you use the produce SD it will provide the correct eigenvalue for every eigenvector.