Show that $e^A$ is positive definite for any Hermitian Matrix $A\in M_{n\times n}(\mathbb{C})$.

I'm not sure that my argument is valid, but by following lemma it seems fine to me. I appreciate any correction and suggestion.

Lemma: If $V$ is an inner product space finitely generated over $\mathbb{C}$. A nomral endomorphism of $\alpha$ is positive definite iff each of its eigenvalues is positive.

Define $\alpha:v\to Av$ where $v\in \mathbb{C}^n$. Since $A$ is Hermitian, so $\alpha$ is normal, and it's orthogonality diagonalizable. Hence there exists an orthonormal basis $B=\{v_1,\dots,v_n\}$ for $\mathbb{C}^n$, composed of eigenvectors of $\alpha.$ Thus, $\phi_{BB}(\alpha)=[a_{ij}]$ is a diagonal matrix where each $a_{ii}$ is eigenvalue of $\alpha$ associated with the eigenvector $v_i$ for all $1\leq i \leq n$. $A$ is similar to $\phi_{BB}(\alpha)$, so there exists a nonsingular matrix $P$ satisfying $A=P^{-1}\phi_{BB}(A)P$.

Since $A$ is hermitian then $e^A$ is Hermitian, so it's normal, and by the properties of an exponential matrix, we can get

$$e^A=P^{-1}\begin{bmatrix} e^{a_{11}}&\dots& 0\\ \vdots&\ddots&\vdots\\ 0&\dots &e^{a_{nn}} \end{bmatrix}P $$ As, all of eigenvalues of $e^A$ is positive, hence it's positive definite.

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    $\begingroup$ Just mention that the eigenvalues are reals. Then everything is alright. $\endgroup$ – HyJu Jan 11 '17 at 4:21

Let $\langle*,*\rangle$ denote the usual inner product on $ \mathbb C^n$ and $\|*\|$ the induced norm on $ \mathbb C^n$. For $x \in \mathbb C^n$ we have, with $B=e^{A/2}$:

$\langle e^Ax,x\rangle=\langle B^2x,x\rangle=\langle Bx,B^{\star}x\rangle=\langle Bx,Bx\rangle=\|Bx\|^2$,

since $B^{\star}=e^{A^{\star}/2}=e^{A/2}=B$.

$B$ is invertible, hence $Bx \ne 0$ if $x \ne 0$. It follows that

$$\langle e^Ax,x\rangle >0$$

if $x \ne 0$.


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