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 $b \in (0,2) \subset \mathbb R$.

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    $\begingroup$ When you say “the answer is $b\in\mathbb{R}$ do you mean its _in_definite for all $b\in\mathbb{R}$? $\endgroup$ – David M. Feb 16 '19 at 15:42
  • $\begingroup$ 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. $\endgroup$ – Hai Feb 16 '19 at 15:47
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    $\begingroup$ Also note that for $b=0$, $q$ reduces to $2x_3^2$ which is clearly not indefinite $\endgroup$ – David M. Feb 16 '19 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.

I think your answer is correct.

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

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