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

hermitian-matriceslinear algebramatrices

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

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

Best Answer

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 $v$.

Related Question