If I have a symmetric matrix $A$ that is negative definite, then spectral decomposition theorem says that there is an orthogonal matrix $Q$(matrix of eigenvectors) such that $A=Q \Lambda Q^T$. Does that mean that $A^2$ cannot be negative definite, since $A^{2}=(Q \Lambda Q^{T})(Q \Lambda Q^{T})$ so $A^{2}=Q \Lambda ^{2} Q^{T}$ and I have the squares of eigenvalues on the diagonal, so all $A^2$ has all positive eigenvalues?
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$\begingroup$It's true that the square of a negative definite matrix is positive definite. You can see this by the spectral decomposition as you pointed out. It's also a general fact that the square of a symmetric matrix must be positive-semidefinite, since$$ v \cdot A^2 v = Av \cdot Av \geq 0$$In your case, the matrix $A$ is also (negative) definite, and hence nonsingular, so you have the strict inequality$$ Av \cdot Av > 0$$showing that $A$ is positive definite.
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