Proving |det(1+A)|^2 >=1 by Diagonalizing iA: Linear Algebra Question

[N00813]

Homework Statement

A is an anti-Hermitian matrix.
Show, by diagonalising iA, that:

|det(1+A)|^2 >=1

Homework Equations

A^H denotes hermitian conjugate of A; A^H = -A\

x = Ox' transforms vector components between 2 basis sets.

The Attempt at a Solution

I know that iA is a Hermitian matrix, so it is diagonalisable.
But I'm not sure what the point of diagonalising iA is.

D = O^(-1)(iA)O
for diagonal matrix of eigenvalues D.

det(iA) = det(D).

I'm floundering around. Any help is appreciated.

 

[pasmith]

Homework Statement

A is an anti-Hermitian matrix.
Show, by diagonalising iA, that:

|det(1+A)|^2 >=1

Homework Equations

A^H denotes hermitian conjugate of A; A^H = -A\

x = Ox' transforms vector components between 2 basis sets.

The Attempt at a Solution

I know that iA is a Hermitian matrix, so it is diagonalisable.
But I'm not sure what the point of diagonalising iA is.

D = O^(-1)(iA)O
for diagonal matrix of eigenvalues D.

Therefore $A = O(-iD)O^{-1}$.

Also the eigenvalues of $iA$ (and hence the entries of $D$) are real.

Hint:
$$O^{-1}(I + A)^2O = O^{-1}(I + 2A + A^2)O \\<br /> = O^{-1}(I + 2A + AOO^{-1}A)O$$
and
$$\det(O^{-1}(I + A)^2O) = \det ((I + A)^2)$$

 

[N00813]

Therefore $A = O(-iD)O^{-1}$.

Also the eigenvalues of $iA$ (and hence the entries of $D$) are real.

Hint:
$$O^{-1}(I + A)^2O = O^{-1}(I + 2A + A^2)O \\<br /> = O^{-1}(I + 2A + AOO^{-1}A)O$$
and
$$\det(O^{-1}(I + A)^2O) = \det ((I + A)^2)$$

I suppose the first equality equals:

$$O^{-1}(I + 2A + AOO^{-1}A)O = (I - 2iD - D^2)$$

(This is the bit I'm a bit wonky on, particularly the 3rd equality. Is that part correct?)
Then $$|\det((I + A)^2)| = |\det(I - 2iD - D^2)| = |\det((I-iD)^2)|=|\det(I^2 + D^2)| >= det(I^2) = 1$$
where

 

[pasmith]

I suppose the first equality equals:

$$O^{-1}(I + 2A + AOO^{-1}A)O = (I - 2iD - D^2)$$

(This is the bit I'm a bit wonky on, particularly the 3rd equality. Is that part correct?)
Then $$|\det((I + A)^2)| = |\det(I - 2iD - D^2)| = |\det((I-iD)^2)|=|\det(I^2 + D^2)| \geq 1$$

If you're not sure that the third equality holds (and it holds only if the eigenvalues of D are real), you should prove it. This will require calculating the quantity you are asked to show is at least 1, at which point you have answered the question.

 

[N00813]

If you're not sure that the third equality holds (and it holds only if the eigenvalues of D are real), you should prove it. This will require calculating the quantity you are asked to show is at least 1, at which point you have answered the question.

I suppose, if the action of the complex conjugate is to change the sign of the i:
$$|\det((I-iD)^2)| = \det((I-iD) (\det(I-iD))^* = \det(I-iD)\det(I+iD)$$

I suppose I can prove $$(\det(I-iD))^* = \det(I+iD)$$ using suffix notation definition for the determinant.

Thanks!