Definiteness of function $Q(x_1, x_2) = x_1^2 - 6 \cdot x_1 \cdot x_2$

Question

I am currently struggling with a simple quadratic form:

$$Q(x_1, x_2) = x_1^2 - 6 \cdot x_1 \cdot x_2$$

It is obvious by completing the square (or by simply look at 3D-graph), that this function is indefinite:

$$Q(x_1, x_2) = (x_1 - 3 \cdot x_2)^2 - 9 \cdot x_2^2$$

Now I want to the exact same analysis with the square matrix A:

$$\mathbf{x}^\top A \mathbf{x}$$

for $A$ I get the matrix:

\begin{bmatrix} 1 & 0 \\ -6 & 0 \\ \end{bmatrix}

If I analyse the matrix with Sylvester criterion (simply by adding row II + 6 * row I) I get the eigenvalues 1 and 0. This means that the matrix is pos. semi-definite?

The matrix is also positive definite, if I apply the Hurwitz criterion to the matrix:

$$k=1: 1 ≥ 0, 0 ≥ 0$$ $$k=2: 1 * 0 - (-6) * 0 ≥ 0$$

What am I doing wrong here? Shouldn't the analysis yield the same outcome?

Answer 1

I'm assuming $A$ is matrix representation of $Q(x_1,x_2)$, if so, then your matrix is wrong. Check here. So $A$ should instead be: \begin{bmatrix} 1 & -3 \\ -3 & 0 \\ \end{bmatrix} Now using Sylvester criterion you could easily prove what you're looking for.