# For what values of $b$ is this quadratic form indefinite?

I'm trying to determine what (real) values of $$b$$ the quadratic form $$q = bx_1^2 + 2bx_2^2 + (9b + 2)x_3^2 − 2bx_1x_2 − 6bx_1x_3 + 4bx_2x_3$$.

I know (using leading principal of minors) that the quadratic form is positive definite for $$b \in (0,2)$$ and positive semi-definite for $$b \in [0,2]$$ and that it cannot be negative definite/semi-definite for any real values of $$b$$. Thus this would indicate that it's indefinite for $$b \in \mathbb R \text{\\} [0,2]$$, however I am told that the answer is $$b \in \mathbb R$$. I can't see why this is correct as the quadratic form will be positive definite for $$b \in (0,2) \subset \mathbb R$$.

• When you say “the answer is $b\in\mathbb{R}$ do you mean its _in_definite for all $b\in\mathbb{R}$? – David M. Feb 16 at 15:42
• Yes. @DavidM. That's why I'm not sure if it's correct since I know the matrix is positive definite for the given interval. – Hai Feb 16 at 15:47
• Also note that for $b=0$, $q$ reduces to $2x_3^2$ which is clearly not indefinite – David M. Feb 16 at 16:08

The quadratic form corresponds to the matrix $$\begin{bmatrix} b&-b&-3b\\ -b&2b&2b\\ -3b&2b&9b+2 \end{bmatrix}$$ Computing the eigenvalues of this matrix for $$b=1$$ gives $$\lambda=\begin{bmatrix} 12.2992\\ 1.65159\\ 0.049229 \end{bmatrix}$$ i.e. the quadratic form is positive definite, whereas for $$b=3$$ we get $$\lambda=\begin{bmatrix} 33.4129\\ 4.64512\\ -0.0579871 \end{bmatrix}$$ i.e. the quadratic form is indefinite.

• $$b (x-y-3z)^2 + b (y-z)^2 + (2-b)z^2$$ where $x=x_1, y=x_2, z=x_3$ – Will Jagy Feb 16 at 19:05