Show that U must Be Unitary

[ognik]

A particular similarity transformation yields:
A^' = UAU^-1
(A^')^{\(\displaystyle \dagger\)} = UA^{\(\displaystyle \dagger\)}U^-1
if the adjoint relationship is preserved (A^{'\(\displaystyle \dagger\)}=A^{\(\displaystyle \dagger\)'}) and det(U)=1, then Show that U must Be Unitary
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I think I've nearly got it as follows, but not sure my arguments are completely valid:
|det(U)|=1 is one of the conditions for U to be Unitary

From A^' = UAU^-1:
$ ({A}^{'})^{\dagger} = (UA{U}^{-1})^\dagger = (A{U}^{-1})^\dagger{U}^{\dagger} = ({U}^{-1})^\dagger {A}^{\dagger} {U}^{\dagger} $

So, from (A^')^{\(\displaystyle \dagger\)} = UA^{\(\displaystyle \dagger\)}U^-1 above:
$ ({U}^{-1})^\dagger{A}^{\dagger}{U}^{\dagger} = U{A}^{\dagger}{U}^{-1} $
Then equating the RH elements, $ {U}^{\dagger}={U}^{-1} $ which makes U unitary...provided we can also show the the LH elements are equal, IE show $ ({U}^{-1})^\dagger = U $
Now if U is Unitary, then $ ({U}^{-1})^\dagger = ({U}^\dagger)^\dagger = U $

This last step I don't think I have argued correctly, how would I improve the argument, or is there a better way of doing this proof? Thanks

 

[jvicens]

for your help!Your reasoning is on the right track, but it can be simplified and made more rigorous. Here's a possible proof:

First, note that since $U$ is a similarity transformation, it must be invertible. Therefore, its adjoint $U^\dagger$ also exists.

Next, we have $(A')^\dagger = (UAU^{-1})^\dagger = (U^{-1})^\dagger A^\dagger U^\dagger$, where we have used the fact that the adjoint of a product is the product of the adjoints in the last step.

Now, since we are given that $(A')^\dagger = A^\dagger$, we can equate the two expressions to get $(U^{-1})^\dagger A^\dagger U^\dagger = A^\dagger$. This holds for any matrix $A$, so we can choose a specific matrix, say $A = I$, the identity matrix. This gives us $(U^{-1})^\dagger U^\dagger = I$.

Finally, we can take the adjoint of both sides to get $(U^\dagger)^\dagger (U^{-1})^\dagger = I^\dagger = I$. But we know that $U$ is unitary if and only if $U^\dagger U = I$, so we can rewrite this as $U^\dagger (U^{-1})^\dagger = I$. Comparing this to our previous equation, we see that $(U^{-1})^\dagger = U^\dagger$, and since $U^\dagger = U^{-1}$, this means that $(U^{-1})^\dagger = U$, as desired.

Note that this proof does not rely on the determinant condition, as the unitarity of $U$ is already built into the definition of a similarity transformation. So we do not need to explicitly show that $|det(U)| = 1$.

I hope this helps!