if Hessian matrix of a function has positive real eigenvalues, then function is positive definite Can we say that if Hessian matrix of a function has positive real eigenvalues, then function is positive definite? Is it true? Would you recommend a document for me to read on this subject?
$V=\alpha x^2+\alpha y^2+ z^2$, I know this function is positive definite for $\alpha >0$. But I wonder if it would be correct to check with its Hessian matrix as I asked above?

Let $H$ be the Hessian matrix of a twice differentiable function. Then $H$ is symmetric. Then, $H$ can be orthogonally diagonalisable $H = \Gamma D \Gamma^\intercal,$ where $D$ is a diagonal matrix with diagonal entries the eigenvalues of $H$ and $\Gamma$ is the orthogonal matrix whose columns are unitary eigenvectors of the $H.$ If $D$ is positive (meaning all eigenvalues are positive), you can trivially define $D^{\frac{1}{2}}$ and then you decompose the Hessian as $\Gamma D^{\frac{1}{2}} D^{\frac{1}{2}} \Gamma^\intercal.$ Let $x \neq 0.$ If you pre and post multiply by $x^\intercal$ and $x$ you find that $x^\intercal H x = \left\| D^{\frac{1}{2}} \Gamma^\intercal x \right\|^2 > 0$ sincc $D^{\frac{1}{2}} \Gamma^\intercal$ is invertible.
