Find a canonical form through orthogonal transformation I need to find a canonical form through orthogonal transformation, the problem is, that the equation given to me doesn't make sense:
$$4x_1^2+4x_2^2+x_3^2-2x_1x_2-2\sqrt{3}x_2x_3 $$
The matrix for this form is:
\begin{pmatrix}
4 & -1 & 0\\
-1 & 4 & -\sqrt{3}\\
0 &-\sqrt{3} & 1
\end{pmatrix}
This matrix are correct? So the equation to find eigenvalues looks like this: $-^3+9^2−20+3=0$. How to solve equation like this? I have such roots  as I have to find transformation matrix
 A: Your quadratic form $$4x_1^2+4x_2^2+x_3^2-2x_1x_2-2\sqrt{3}x_2x_3 $$ can be written as
$$v^{\top}Qv
=
\left(\begin{array}{c}
x_1\\x_2\\x_3
\end{array}
\right)^{\top}\left(
\begin{array}{ccc}
 4 & -1 & 0 \\
 -1 & 4 & -\sqrt{3} \\
 0 & -\sqrt{3} & 1 \\
\end{array}
\right)
\left(\begin{array}{c}
x_1\\x_2\\x_3
\end{array}
\right).$$
If we introduce a basis change in order to express this at its
easier viewpoint one choose the $Q$ eigenvectors.
The next matrix shows them in the columns
$$S=\left(
\begin{array}{ccc}
 0.55415 & 0.82474 & 0.11279 \\
 -0.77451 & 0.46119 & 0.43294 \\
 0.30505 & -0.32727 & 0.89434 \\
\end{array}
\right).$$
Now, manipulate as
\begin{eqnarray*}
v^{\top}Qv&=&(SS^{-1}v)^{\top}QSS^{-1}v\\
&=&(S^{-1}v)^{\top}\ (S^{\top}QS)\ S^{-1}v\\
\end{eqnarray*}
to achieve the effect of writing your quadratic form into the form
$$v^{\top}Qv=q_1y_1^2+q_2y_2^2+q_3y_3^2,$$
where
$$\left(\begin{array}{c}
y_1\\y_2\\y_3
\end{array}
\right)=S^{-1}v$$
are the components of $v$ in the new basis.
It easy to calculate that $S^{\top}QS$ would be the diagonal matrix
$$\left(
\begin{array}{ccc}
 q_1 & 0 & 0 \\
 0 & q_2 & 0 \\
 0 & 0 & q_3 \\
\end{array}
\right).$$
Observe that $S^{-1}QS$ gives the diagonalization of $Q$
which has the eigenvalues in the principal diagonal as
$$\left(
\begin{array}{ccc}
 5.3977 & 0 & 0 \\
 0 & 3.4408 & 0 \\
 0 & 0 & 0.16153 \\
\end{array}
\right).$$
If instead of choosing the eigenvectors suggested by Wolfram-Alpha
you use a normalization of them then the matrix $S$ would be orthogonal.
