I am following the solution of a problem which has the following matrix $$\begin{bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$$ along with $dz = -dx - dy$ and also corresponding quadratic form is $2dx^2 + 2dy^2$. Looking at the matrix there is clearly an eigenvalue of $0$, yet the problem I am following says using the identity $dz = -dx - dy$ it is possible to "eliminate" $dz$ and show that the quadratic form is actually positive. How is $\lambda = 0$ discarded?
2026-04-04 12:06:49.1775304409
Positive Definite Quadratic Form
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If $dz=-dx-dy$ then $[dz]^TAdz=(-1,-1,0)A(-1,-1,0)^T$ and because $A$ is positive definite with respect to $x$ and $y$ its apparent positive semi-definiteness is actually properly seen as positive definite. You're eliminating $dz$ as a variable from your space, which amounts to expressing it in the basis $dx,dy$ as that identity already does. If it had a $-1$ in the lower-right corner, then this simplification would be impossible (and the identity would be wrong because of the quadratic form associated with the matrix).