Let $A$ be a Hermitian matrix, prove that $v^*Av >0$ for all $0 \not = v \in \Bbb C^n$ $\iff$ all the eigenvalues of $A$ are greater than zero

Question

Let $A \in \Bbb C^{n \times n}$ be a Hermitian matrix, prove that $v^*Av >0$ for all $0 \not = v \in \Bbb C^n$ $\iff$ all the eigenvalues of $A$ are greater than zero

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I started this by stating everything I know about Hermitian matrices:

1. $A=A^*$
2. all the eigenvalues are real so $\lambda \in \Bbb R$
3. There exists a spectral decomposition for $A$
4. For different eigenvalues $\lambda_1 \dots \lambda_r$ and eigenspaces $V_{\lambda_1} \dots V_{\lambda_r}$ we have $\Bbb C^n = V_{\lambda_1} \oplus \dots \oplus V_{\lambda_r}$ and also $V_{\lambda_i} \perp V_{\lambda_j}$ for all $i \not =j$
5. It is unitarily diagonalizable
6. $A$ is also a normal matrix

So start from the first side assume $v^*Av>0$ we know that $Av= \lambda v$ so $v^*Av=v^* \lambda v= \lambda v^* v$ I was not too sure here but I think since it has a spectral decomposition then it also has an orthonormal basis(?) so $v^*v=1$ then we get $\lambda >0$

for the second side assume $\lambda >0$ we know it has a spectral decomposition because it is Hermitian matrix so $A= \lambda_1 \hat v_1 \hat v_1^* + \dots + \lambda_n \hat v_n \hat v_n^*$

because of the spectral decomposition $\{v_1 \dots v_n\}$ are orthonormal basis for $\Bbb C^n$ that is made of eigenvectors of $A$ But I got stuck even after substituting $A$ in $v^*Av$ I could not see how to continue

Another question that is not related to this question but As for unitary matrices, what can I conclude if it is given that I have a unitary matrix?(Like in the first part of the question I had 6 points about Hermitian matrix)

Thanks for any tips and help! hopefully the translations are understandable

Edit:

Thanks for the link to the similar question , it is indeed similar but I am trying to solve it with spectral theorem and I am trying to see if my approach is actually correct for the first side of the proof about the orthonormal basis . the question provided does not answer that and also how to solve with spectral decomposition the other side

Answer 1

Your basic idea for one direction of the proof is correct. You didn't phrase it as an argument/proof, but here is one way of doing so.

Suppose that $A$ is a Hermitian and that $v^*Av > 0$ holds for all non-zero $v$. Let $\lambda$ be any eigenvalue of $A$, and let $v$ be an associated eigenvector. We have $$ 0 < v^*Av = v^*(\lambda v) = \lambda v^*v = \lambda \|v\|. $$ It follows that $\lambda > 0$. Thus, every eigenvalue of $A$ must be positive

For the other direction of the proof: if you substitute your expression of $A$ into $v^*Av$, you can rewrite the result in the form $$ v^*Av = \lambda_1 |v^*\hat v_1|^2 + \cdots + \lambda_n |v^*\hat v_n|^2. $$ From here, it's easy to see that if $\lambda_1,\dots,\lambda_n$ are positive, then it must hold that $v^*Av > 0$ for all non-zero $v$.