Inner product, positive-definiteness

Question

I want to define for which $a,b,c \in \mathbb{R}$ $$(X,Y) = X^T \begin{pmatrix} b+6 & 0 & 11 \\ 0 & 6 & a \\ 11 & -2 & 1 \end{pmatrix} Y + cy_1^2$$ is an inner product of $\mathbb{R}^3$. ($y_1$ first coord. of $Y$).

I have shown that $a = -2, c = 0$ by just controlling symmetry and bilinearity of $( \cdot, \cdot)$. For $b$, I am having trouble actually proving the boundaries of $b$. Using positive-definiteness, I have shown that $b > -6$ but I cannot prove that these are the only values of which $( \cdot, \cdot)$ is an inner product. How would I go about actually proving the boundaries of $b$?

Thank you.

Answer 1

Sylvester's criterion says the matrix is positive definite if and only if all its principal minors are positive. That is, calculate the determinant of the $1\times1$ matrix in the upper left corner, the $2\times2$ in the upper left corner, and the full $3\times3$; if those three numbers are all positive, you win.

Answer 2

An inner product (over a real vector space) has to satisfy three conditions:

1. Positive definiteness
2. Symmetry
3. Bilinearity

As you already stated, symmetry and bilinearity give you $a = -2$ and $c=0$.

Positive definiteness of the inner product is equivalent to positive definiteness of its 'generating' matrix which in turn is equivalent to all eigenvalues being strictly positive. Hence, it suffices to calculate all three eigenvalues of your matrix (with dependence on $b$) and check for which $b$'s they are positive.

If you need help with these eigenvalues, leave a comment.