Hermitian matrices and unitary similarity transformations

[ShayanJ]

I tried to prove that a hermitian matrix remains hermitian under a unitary similarity transformation.I just could do it to he point shown below.Any ideas?

[itex] [ ( U A U ^ {\dagger}) B ] ^ {\dagger} = B ^ {\dagger} (U A U ^ {\dagger}) ^ {\dagger} = B (U A^ {\dagger} U ^ {\dagger}) [/itex]

thanks

 

[tiny-tim]

Hi Shyan!

What's B doing there? ​

 

[ShayanJ]

Isn't B the hermitian matrix which should remain hermitian?

 

[tiny-tim]

But A is the matrix which undergoes the similarity transformation.

 

[blue_raver22]

It is true that a Hermitian matrix remains Hermitian under a unitary similarity transformation. This is because unitary matrices preserve the Hermitian structure of a matrix, meaning that the Hermitian property is not affected by a unitary transformation.

To prove this, we can use the fact that a unitary matrix U satisfies U^{\dagger} = U^{-1}. Therefore, we can rewrite the expression as:

[(UAU^{\dagger})B]^{\dagger} = (U^{\dagger})^{\dagger}A^{\dagger}U^{\dagger}B^{\dagger} = UAU^{\dagger}B^{\dagger}

Since A is Hermitian, we know that A^{\dagger} = A. And since B is an arbitrary matrix, we can switch the order of multiplication without affecting the result. Therefore, we can rewrite the expression as:

UAU^{\dagger}B^{\dagger} = UAB^{\dagger} = UBA^{\dagger} = (UBA^{\dagger})^{\dagger}

This shows that the resulting matrix after the unitary similarity transformation is also Hermitian, thus proving that a Hermitian matrix remains Hermitian under a unitary similarity transformation.

I hope this helps and provides some further insight into the topic. If you have any other questions or concerns, please feel free to ask. Keep up the good work in exploring the properties of Hermitian matrices and unitary transformations!