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$S$ is real symmetric. $T$ is real skew-symmetric. I have shown that $T\pm iS$ is skew-Hermitian. I am further asked to show that $U = (I+T+iS)(I-T-iS)^{-1}$ is unitary.

Denoting by $^\dagger$ the conjugate transpose, I have that
$U^\dagger = [(I+T+iS)(I-T-iS)^{-1}]^\dagger$
$= ((I-T-iS)^\dagger)^{-1}(I+T+iS)^\dagger$
$= (I+T+iS)^{-1}(I-T-iS)$

But then the products $UU^\dagger$ and $U^\dagger U$ have the factors in the wrong order to cancel out to make identity. I must be missing something obvious here; I'd appreciate a steer!

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For any $A$, $(I+A)$ commutes with $(I-A)$, and hence with $(I-A)^{-1}$ if it exists. –  Chris Eagle Jan 5 '11 at 19:25
    
Thanks Chris, that makes it easy. I don't think I've seen that simple (but useful-looking) result until now! I have proved it using index notation; is there a nicer way? –  user5426 Jan 5 '11 at 19:50
    
If $A$ commutes with $B$, then $AB=BA$. If further $B$ is invertible, then $B^{-1}(AB)B^{-1}=B^{-1}(BA)B^{-1}$, and hence $B^{-1}A=AB^{-1}$. –  Chris Eagle Jan 5 '11 at 19:58
    
Sorted, thanks very much indeed. –  user5426 Jan 6 '11 at 1:09

1 Answer 1

I use the hints in the comments of Chris Eagle to give this old question an answer.

  1. If $AB = BA$ for any square matrices $A,B$ and $A$ is invertible, then $A^{-1} B = BA^{-1}$. (Proof: Multiply $AB = BA$ from both sides with $A^{-1}$.)

  2. For any square matrix $A$, the matrices $I + A$ and $I - A$ commute (Proof: $(I + A)(I - A) = I\cdot I + A\cdot I - A\cdot I - A\cdot A = I - A^2$ and in the same way $(I - A)(I + A) = I - A^2$, too.

  3. By the calculations carried out in the question, we have $$ UU^\dagger = (I + (T + iS))(I - (T + iS))^{-1}(I + (T + iS))^{-1}(I - (T + iS)). $$ By 1. and 2., the first two factors do commute (in fact, all four factors are pairwise commutable). So $$ UU^\dagger = (I - (T + iS))^{-1}\underbrace{(I + (T + iS))(I + (T + iS))^{-1}}_{=I}(I - (T + iS)) = (I - (T + iS))^{-1}(I - (T + iS)) = I$$ and $U$ is unitary.

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