Matrix Proof using Unitary operators

[andre220]

Homework Statement

Show that if two square matrices of the same rank are related by unitary transformation $\hat{A}=\hat{U}^\dagger\hat{B}\hat{U}$ then their traces and determinants are the same.

Homework Equations

$Tr(\hat{A}) =\sum\limits_{k=0}^{n}a_{kk}$
$\hat{U}^\dagger\hat{U} = 1$

The Attempt at a Solution

Ok so I have no idea where to start with this, my first thought is to expand the RHS of the transformation:

$$=\left(u_{ij}\right)^\dagger b_{ij} u_{ij} = \left(u_{ji}b_{ji}^*\right)u_{ij}$$

But I am not sure if this right or where to go from there.

 

[Orodruin]

I think parts of your post are missing. As I understand it, you want to prove that the trace is invariant under unitary transformation. What properties of the trace are you familiar with? Can you write out the expression for the components of A?

 

[andre220]

You're right I was missing two parts of it, I edited the initial post. It is correct now. For component of A, I am not quite what you mean, something like this?: $$A = \begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n} \end{pmatrix}$$

And as for the properties of the trace: I'm familiar with most basic properties (I apologize, my einstein summation indices were not correct in the first post): i.e. $Tr(AB) = Tr(BA)$, $Tr(BAB^{-1}) = Tr(A)$, for example.

 

[WWGD]

I think proving the weaker result that similar matrices have the same trace, determinant proves what you want, since being unitarily-equivalent/similar is stronger than just being similar.

 

[Dick]

I guess I don't understand why you aren't applying the property of trace to your problem. Pretty much the same for determinant. No need to look at components.

 

[Orodruin]

As Dick said, use the properties of the trace and determinant and the problem should be trivial. The determinant property you will need is ${\rm det}(CD) = {\rm det}(C){\rm det}(D)$.

 

[andre220]

Yeah after he said that I looked at it an yeah I was making too much of it: essentially it is just:

$\mathrm{Tr}(\hat{A})=\mathrm{Tr}(\hat{U}^\dagger\hat{B}\hat{U}) = \mathrm{Tr}(\hat{B}\hat{U}^\dagger\hat{U}) = \mathrm{Tr}(\hat{B}\hat{1}) = \mathrm{Tr}(\hat{B})$

and then it would almost identical for the determinant. Thank you for your help.

 

[Fredrik]

$Tr(\hat{A}) =\sum\limits_{k=0}^{n}a_{kk}$

$$=\left(u_{ij}\right)^\dagger b_{ij} u_{ij} = \left(u_{ji}b_{ji}^*\right)u_{ij}$$

The definition of matrix multiplication is $(AB)_{ij}=A_{ik}B_{kj}$ (with summation over repeated indices). So
\begin{align}
&\operatorname{Tr A}=A_{kk}=(U^\dagger BU)_{kk}=(U^\dagger)_{ki}(BU)_{ik}=(U^\dagger)_{ki}B_{ij}U_{jk} =B_{ij}U_{jk}(U^\dagger)_{ki} =B_{ij}(UU^\dagger)_{ji}\\ &=B_{ij}\delta_{ji}=B_{ii}=\operatorname{Tr} B.
\end{align}
This is how you prove that the trace has that nice property. The proof that determinants are equally nice is more complicated, because the definition of "determinant" is more complicated than the definition of "trace".

 

[andre220]

Yeah after he said that I looked at it an yeah I was making too much of it: essentially it is just:

$\mathrm{Tr}(\hat{A})=\mathrm{Tr}(\hat{U}^\dagger\hat{B}\hat{U}) = \mathrm{Tr}(\hat{B}\hat{U}^\dagger\hat{U}) = \mathrm{Tr}(\hat{B}\hat{1}) = \mathrm{Tr}(\hat{B})$

and then it would almost identical for the determinant. Thank you for your help

 

[andre220]

Ahh okay that makes sense, I'm still getting the hang of the sum notation.

 

[andre220]

THank you.