Question

Which is the easiest way to prove that "If A is a unitary matrix and \( A^{-1} \) is also unitary matrix"

Answer 1

If A is unitary matrix, then
\[ A^{-1} = A^\dagger \]
where \( A^\dagger \) is the conjugate transpose.

As \[ (A^\dagger)^\dagger = A \]
you should have your result.

Answer 2

nice dagger!

Answer 3

duh it was sooo easy, thanks James.

Answer 4

now tell me how could we prove that |det (A)|=1

Answer 5

\[(A^\dagger)^\dagger A^\dagger= AA^\dagger=I\]just testing

Answer 6

Ah, that's straight forward too:
\[ \det(A^\dagger) = (\det{A})^* \]
where * means conjugate

Now as \( AA^\dagger = I \)
\[ 1 = \det I = \det(AA^\dagger) = \det A . (\det A)^* \]
\[ \implies |\det A|^2 = 1 \implies |\det A| = 1 \]

Answer 7

pretty!

Answer 8

whats the * mean there james?

Answer 9

oops

Answer 10

complex conjugate. Hence ...
\[ \det A^\dagger = \det (A^T)^* = (\det A^T)^* = (\det A)^* \]

Answer 11

nvm i didn't read that line

Answer 12

ok perfect

Answer 13

hmmm i am sure it is right, but why does
\[|\det A|^2 = 1 \implies |\det A| = 1\]?

Answer 14

because |det A| has to be a non-negative number

Answer 15

In other words, |det A|^2 = 1 implies |det A| = +1 or -1. But it can't equal -1.

Answer 16

oh right.

Answer 17

I still can't follow how \( \det A . (\det A)^* = |\det A| ^2 \) ?

Answer 18

In genreal, for any complex number z,
\[ zz^* = |z|^2 \]

Answer 19

Aw, right! and your first result I think that is consistent to this : http://en.wikipedia.org/wiki/Determinant#Square_matrices_over_commutative_rings_and_abstract_properties ?

Answer 20

That's a messy one though, do you know any easy proof of that property?

Answer 21

which property in particular?

Do you mean that
\[ A^\dagger = A^{-1} \]
or something else?

Answer 22

That can be deduced easily from the definition of unitary matrices, I was wondering about \( \det(A^\dagger) = (\det{A})^* \)

In words, the determinant of the complex conjugate of a complex matrix (which is also the determinant of its conjugate transpose) is the complex conjugate of its determinant, right ?

Answer 23

oh, that's what I wrote out above.

dagger = transpose and conjugate. Now the det of a transpose of a matrix is equal to the det of the original matrix. So det of A-dagger = the conjugate of det A.

Answer 24

yes yes, but how to prove it ?

Answer 25

Carefully,
\[ \det[ A^\dagger ] = \det [ (A^T)* ] \ \ \ \hbox{by definition} \]
\[ = (\det[ A^T ])^* \]
because the conjugate of a sum of product of complex numbers is the sum of the product of the conjugate of each number, because (z+w)* = z* + w* and (zw)* = z*w*
\[ = (\det A)^* \ \ \ \ \hbox{ because } \det A^T = \det A \ \hbox{ for all square matrices } \]

Answer 26

Cool! should have reviewed complex numbers once, Thanks james.