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i am wondering if it is necessary to find the eigenvalues of the Hessian matrix in dimensions > 2 when we are looking for the extrema of a function in the case that one wants to resolve this problem through the way of considering the definitiness of the Hessian.

In other words how do i prove if the Hessian is negative definite if i wouldn't like to go through the eigenvalue problem ?

Thanks for the comment.

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What you probably want is Sylvester's criterion: a Hermitian (and hence symmetric, which is sufficient for the Hessian) matrix $$ \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots \\ a_{21} & a_{22} & a_{23} & \cdots \\ a_{31} & a_{32} & a_{33} & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix} $$ is positive-definite if and only if the determinants of the principal minors, $$ a_{11}, \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}, \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}, \dotsc $$ are all positive. This is obviously much easier to calculate than doing all the eigenvalues.

(See also this question for a simpler proof than Wikipedia's)

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