Transforming the quadratic form $x_1x_2 + x_2x_3 + x_3x_1$ to its normal form. I need to transform $x_1x_2 + x_2x_3 + x_3x_1$ to its normal form described by the inertia theorem. How can I find the transformation? I am allowed to complete squares and perform rotations to find the linear change of coordinates. I know how to do a rotation for quadratic forms over $\mathbb{R}^2$ but not $\mathbb{R}^3$. How would I go about manipulating this expression?
 A: The matrix of the quadratic form for the given basis $x_1, x_2, x_3$ is
$$
A = \frac 12\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}.
$$
For simplicity, I will diagonalize $2A$ first. Eigenvalues of $2A$ can be found by solving
\begin{align}
\lambda^3 - 3 \lambda - 2 & = 0 \\
(\lambda + 1)^2(\lambda - 2) & = 0.
\end{align}
For $\lambda = -1$, the corresponding eigenvector must satisfy
$$
\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
u_1\\
u_2\\
u_3
\end{pmatrix}=
\begin{pmatrix}
-u_1\\
-u_2\\
-u_3
\end{pmatrix} \Longleftrightarrow
\begin{aligned}
u_1 + u_2 + u_3 = 0.
\end{aligned}
$$
We can pick two arbitrary independent orthogonal solutions. I will pick
$$
u = \begin{pmatrix}
1/\sqrt 6 \\
1/\sqrt 6 \\
-2/\sqrt 6
\end{pmatrix} \quad \text{ and } \quad
v = \begin{pmatrix}
1/\sqrt 2 \\
-1/\sqrt 2 \\
0
\end{pmatrix}.
$$
For $\lambda = 2$, we solve for $w$ from
$$
\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
w_1\\
w_2\\
w_3\end{pmatrix}
=
\begin{pmatrix}
2w_1\\
2w_2\\
2w_3
\end{pmatrix} \Longleftrightarrow
\begin{aligned}
-2w_1 + w_2 + w_3 & = 0\\
w_1 - 2w_2 + w_3 & = 0\\
\end{aligned}.
$$
It is easy to see that $w = \begin{pmatrix}1/\sqrt 3\\1/\sqrt 3\\1/\sqrt 3\end{pmatrix}$ is a solution.
Therefore,
$$
2A = \begin{pmatrix}
1/\sqrt 6 & 1/\sqrt 2 & 1/\sqrt 3\\
1/\sqrt 6 & -1/\sqrt 2 & 1/\sqrt 3\\
-2/\sqrt 6 & 0 & 1/\sqrt 3
\end{pmatrix}
\begin{pmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 2
\end{pmatrix}
\begin{pmatrix}
1/\sqrt 6 & 1/\sqrt 2 & 1/\sqrt 3\\
1/\sqrt 6 & -1/\sqrt 2 & 1/\sqrt 3\\
-2/\sqrt 6 & 0 & 1/\sqrt 3
\end{pmatrix}^T.
$$
The factor $2$ (or its reciprocal) can be pushed to the middle matrix of eigenvalues. This does not change the eigenvectors. The matrix of eigenvectors is the change of coordinates that you want.
A: Hint
$f = \begin{pmatrix}
x, y, z\\
\end{pmatrix}\begin{pmatrix}
0, 1, 0\\
0, 0, 1\\
1, 0, 0\\
\end{pmatrix}\begin{pmatrix}
x\\ 
y\\ 
z\\
\end{pmatrix}$
And you need to represent the matrix A in the form: $S^TDS$, where  conjugate transpose of S: $S^* = S^{-1}$. So the matrix S gives you necessary mapping to get the normal form. And the coefficients of D gives you coefficients of the normal form.
P.S. There will be complex roots. 
