How can I prove the inverse of a unitary matrix is unitary?

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SUMMARY

The discussion centers on proving that the inverse of a unitary matrix is also unitary. It establishes that if matrix A is unitary, then A-1 is unitary by demonstrating that AHA = I leads to A-1(AH)-1 = I. Additionally, it confirms that if A is normal and invertible, A-1 remains normal by using the property that (AH)-1 = (A-1)H. The proofs leverage fundamental properties of matrix operations and the definitions of unitary and normal matrices.

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  • Familiarity with normal matrices and their definitions
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chuy52506
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A is a matrix in the complex field
Suppose A is unitary show that A-1 is unitary.

Suppose A is normal and invertible, show A-1 is normal.

Can i prove the first one just by:
AAT=I
then AT=A-1

Then
I=A-1(AT)-1
So,
I=A-1(A-1)T


I have no idea in how to start the second one? please help
 
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I assume that when you write AT you mean the Hermitian conjugate of A. For the sake of clarity, I'll use AH:=(A*)t to mean the Hermitian conjugate.

Then yeah, your first proof is ok provided you're happy with the fact that
(AH)-1 = (A-1)H ----- (**)
which isn't that hard to prove.

Here's a slightly different proof:
A unitary <==> AHA = I , ( where I = unit matrix)
take inverse of both sides (can do because it's assumed invertible)
A-1(AH)-1 = I
multiply on left by A and right by A-1
A A-1(AH)-1A-1 = (AH)-1A-1 = I
use the result (**) and you're done.

Basically the same proof works for normal matrices.
A normal <==> AH A = A AH
Take the inverse
A-1 (AH)-1 = (AH)-1 A-1
Use the result (**) and you see that A-1 must be normal
A-1 (A-1)H = (A-1)H A-1 .
 

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