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I'm trying to prove that:

$A$ positive definite Hermitian $\iff$ $\exists P \in GL(E)$ s.t. $A=P^*P$.

$GL(E)$ is the set of all invertible matrices.

Proof:

$\Leftarrow:$

Suppose $\exists P \in GL(E)$ s.t. $A=P^*P$

To show: $A=A^*$. We know $A=P^*P.$ And $A^*=\overline{(P^*P)^T} = \overline{P^TP^{*T}} = P^*P \Rightarrow A=A^* \Rightarrow A$ is Hermitian $\Rightarrow \lambda \in \mathbb{R}$(eigenvalues) $\Rightarrow x^TAx =0 \iff x=0 \Rightarrow A$ is positive definite Hermitian.

$\Rightarrow:$

How can I show this? And is my proof so far correct?

karnan
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1 Answers1

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You're missing a transpose

$$ \overline{(P^* P)^T} = \overline{P^T P^{*T}}\\ = \overline{P^T} \overline{P^{*T}}\\ = P^* P $$.

You didn't finish the proof for positivity. You only gave the equivalent condition that you need to prove.

$$ x^* A x = x^* P^* P x = \mid Px \mid^2 == 0\\ $$

implies $Px = 0$ which then implies $x=0$ because $P \in GL(E)$.

For the other direction, you know that $A = Q^{-1} \Lambda Q$ with $\Lambda$ a diagonal matrix with the positive eigenvalues. $Q$ being the change of basis. $\Lambda$ has a positive square root because you can take positive square roots of all the diagonal entries. Call that $D$. It is real, diagonal and invertible. Then $A=Q^{-1} D^* D Q$. Now use what you know about $Q^{-1}$ and $Q^*$.

AHusain
  • 5,183
  • $x^P^Px = (Px)^(Px)$ Why can we rewrite this as $|Px|^2$? And for the last part, since $Q^{-1}=Q^$ we have that $A=(DQ)^DQ$ and now I have to set $DQ=P$. Is it $Q^{-1} = Q^$ since $A$ is Hermitian? – karnan May 05 '20 at 08:53
  • Look at the definition of norm when vector has complex entries. Take $A^=A$ and see if you can use that to prove that Q is unitary ($Q^ = Q^{-1}$). – AHusain May 05 '20 at 17:44