##Small introduction##
Let $\mathbf{x},\mathbf{y}\in\mathbb{R}^n$ be a vectors and $\mathcal{F}: \mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}$ is bilinear form, i.e. map which is linear with bouth arguments. Because of linearity of map $\mathcal{F}$ it is possible to represent it using $n\times n$ matrix $F$ in a such way:
$$
\mathcal{F}(\mathbf{x},\mathbf{y}) = \mathbf{x}^{\top}F\mathbf{y}.
$$
The map $\mathcal{Q}: \mathbb{R}^n\to\mathbb{R}$ giving by formula $\mathcal{Q}(\mathbf{x}) = \mathcal{F}(\mathbf{x},\mathbf{x})$ is called quadratic form and this is what Silvester's law of inertia applied to. There are two ways to represent quadratic form:
- using matrix: $\mathcal{Q}(\mathbf{x}) = \mathbf{x}^{\top}A\mathbf{x}$;
- "as is": $$\mathcal{Q}(\mathbf{x}) = \sum_{i,j=1}^{n}a_i^jx_ix_j,$$ where $a_i^j$ is element of $i$-th row and $j$-th column of matrix $A$ and $x_i$ is $i$-th component of vecotr $\mathbb{x}$.
We may consider only symmentric matrices because if matrix $A$ is not symmetric we may substitude it with symmetric matrix $A'$ such that ${a'}_i^j = {a'}_j^i = (a_i^j + a_j^i)/2$.
##Law of inertia##
Quadratic form is defined positive if it is positive for all arguments. It's easy to determine if form $\mathcal{Q}$ positive defined or not if it is diagonal, i.e. if
$$
\mathcal{Q}(\mathbf{x}) = \sum_{i=1}^{n}a_i^ix_i^2.
$$
Indeed, it's obvious that $\mathcal{Q}$ is positive defined iff each number $a_i^i$ is positive. What should we do if form $\mathcal{Q}$ is not diagonal? If we consider form as a matrix the diagonal form will be represented with diagonal matrix. There is a standard procedure to convert arbitrary symmetric matrix $A$ to diagonal form $V^{-1}DV$, where $D$ is diagonal (there is a [theorem][1] that matrix has diagonal form in basis of its eigenvectors and if the matrix is symmetric that diagonal matrix is real). So if we take arbitrary symmetric matrix $A = V^{-1}DV$ and substitute this to the representation of $\mathcal{Q}(\mathbb{x})$ we will get $\mathcal{Q}(\mathbf{x}) = \mathbf{x}^{\top}V^{-1}DV\mathbf{x}$. We always may achive the matrix $V$ to be orthogonal, i.e. $V^{-1} = V^{\top}$, then
$$
\mathcal{Q}(\mathbf{x}) = \mathbf{x}^{\top}V^{\top}DV\mathbf{x} = (V\mathbf{x})^{\top}DV\mathbf{x} = \widetilde{\mathcal{Q}}(V\mathbb{x})
$$
where the $\widetilde{\mathcal{Q}}$ form is diagonal and it is easy to determine its positive definition. Actually we may use any matrix $V$ such that $V^{\top}DV = A$.
The qeustion is: will be form $\mathcal{Q}$ be definite psotive iff form $\widetilde{\mathcal{Q}}$ is definite positive? The answer is given by Silvester's law of inertia and it is positive. Moreover, count of negative, positive coefficients and zeros (so-called inertia indices) of diagonal form $\widetilde{\mathcal{Q}}$ does not depend on the way form $\mathcal{Q}$ was diagonalized.
##Example##
Now we convert the form giving by the matrix
$$
A =
\begin{pmatrix}
-1 & -1 & -1 \\
-1 & 1 & 0 \\
-1 & 0 & 1
\end{pmatrix}
$$
from the question to the diagonal form by Lagrange method.
Consider $\mathbb{x} = (x,y,z)^{\top}$, then
$$
\begin{align}
\mathcal{Q}(\mathbb{x}) = -x^2 - 2xy + y^2 -2xz + z^2 = \\
-(x^2 + 2xy + 2xz) + y^2 + z^2 =\\
-\left((x+y+z)^2 - (y+z)^2\right) + y^2 + z^2 =\\
-(x+y+z)^2 + 2y^2 + 2yz + 2z^2 = \\
-(x+y+z)^2 + 2\left(\left(y + \frac{z}{2}\right)^2 - \frac{z^2}{2}\right) + 2z^2 = \\ -(x+y+z)^2 + 2\left(y + \frac{z}{2}\right)^2 + \frac{3}{2}z^2.
\end{align}
$$
If we replace arguments $x,y$ and $z$ with $x+y+z$, $y + z/2$ and $z$ then we will take the diagonal form with the same inertia indicies (according to Silvester's theorem). Replacing of arguments in the matrix form is substitution $\mathbf{x}$ with $V\mathbf{x}$, where
$$
V =
\begin{pmatrix}
1 & 1 & 1 \\
0 & 1 & 1/2 \\
0 & 0 & 1
\end{pmatrix}.
$$
It's easy to see that $V^{\top}DV = A$, where $D = \mathrm{diag}(-1, 2, 3/2)$.
We now can see that there is one negative coefficient and two positive ones, so inertia indicies is $n_{-} = 1$ and $n_{+} = 2$ and this form does not definite positive.
[1]: https://en.wikipedia.org/wiki/Spectral_theorem
[2]: http://math.stackexchange.com/a/1817916/346332
