# Linear Algebra question

1. Mar 23, 2014

### N00813

1. The problem statement, all variables and given/known data
A is an anti-Hermitian matrix.
Show, by diagonalising iA, that:

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

2. Relevant equations

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

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

3. 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.

2. Mar 23, 2014

### pasmith

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

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

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

3. Mar 23, 2014

### N00813

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

Last edited: Mar 23, 2014
4. Mar 23, 2014

### pasmith

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.

Last edited: Mar 23, 2014
5. Mar 23, 2014

### N00813

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!