# If $A$ is negative definite, can $A$ squared be negative definite?

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?

• That's right. You can also see it directly: $A$ is negative definite if $v \cdot Av < 0$ for all nonzero $v$. Using the fact that $A$ is symmetric, we get $v \cdot A^2 v = Av \cdot Av \geq 0$. You can get that $Av \cdot Av > 0$ by knowing that $A$ is nonsingular, since $A$ is (negative) definite. Commented Oct 15, 2018 at 12:25
• @Joppy You should post this as an answer so the question won't remain open forever... Commented Oct 15, 2018 at 12:32
• @Joppy I would certainly give that answer a $+1$. Commented Oct 15, 2018 at 13:02

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.