Once again, I'm stuck into a Linear Algebra question. I'm worried since I just could answer an implication of this theorem. Some help would be nice.

Theorem: Let $A \in M_n(\mathbb{C})$. Prove the next are equivalent:

a) $A$ is positive definite

b) $A$ is hermitian and all of its eigenvalues are positive.

c) $A$ has a positive definite square root, i.e. $\exists B$ positive definite such that $B^2=A$.

d) $A$ has a hermitian square root.

e) There exists a matrix $C$ such that $A=C^* C $.

The only implication I've proved is b => a. Thanks a lot in advance.

EDIT: I solved: a <=> b <=> c, just looking for c <=> d, d <=> e, e <=> a.

EDIT 2: I think c<=> d is a direct consequence of a <=> b <=> c, so it's only necessary to prove d <=> e and e <=> a.

EDIT 3: I've proved d <=> e, just need e <=> a. Thanks P. Hagemann :)


I have tried the first two implications a) => b) and b) => c), but I am not sure about the first one.

a) => b): $0<$($e_{i}$+$e_{j}$)$^H$A($e_{i}$+$e_{j}$)=$a_{ii}$+$a_{jj}$+$a_{ij}$+$a_{ji}$ Since $a_{ii}>0$ and $a_{jj}>0$, we conclude that $a_{ij}+a_{ji}$ is real. Similarly, $0<(e_{i}+ie_{j})^HA(e_{i}+ie_{j}$)=$a_{ii}$+$a_{jj}$-$i$$a_{ji}$+$i$$a_{ij}$. Hence, $i$($a_{ij}-a_{ji}$) is real, therefore $a_{ij}$-$a_{ji}$ is imaginary only. With those two properties($a_{ij}+a_{ji}$ is real and $a_{ij}-a_{ji}$ is imaginary), we conclude that $\overline{a_{ij}}$=$a_{ji}$.

Since A is hermitian, A is normal and we can use spectral theorem. Therefore, there exists an unitary matrix Q such that
$Q A Q^H$=$diag(\lambda_{1},..,\lambda_{n})$.

With $0<x^HAx=x^HQ^Hdiag(\lambda_{1},..,\lambda_{n}) Qx$ follows that all eigenvalues have to be positive, if you choose x in the right way.

b)=>c) With spectral theorem again, it follows that there exists an unitary matrix Q such that
$Q A Q^H$=$diag(\lambda_{1},..,\lambda_{n})$ or $ A $=$Q^H diag(\lambda_{1},..,\lambda_{n}) Q$ . Now choose $D=diag(\sqrt{\lambda_{1}},...,\sqrt{\lambda_{n}})$ and take a look at $(Q^H DQ)^2$.


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