Before continuing, let me add the caution that a symmetric matrix can violate your rules and still be positive definite, give me a minute to check the eigenvalues
$$ H \; = \;
\left( \begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3
\end{array}
\right) .
$$
This is positive by Sylvester's Law of Inertia,
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
A proof is given here as a consequence of Gershgorin's circle theorem. For additional information, see
http://en.wikipedia.org/wiki/Diagonally_dominant_matrix and http://mathworld.wolfram.com/DiagonallyDominantMatrix.html
or just Google "diagonally dominant symmetric"
Later methodology, amounting to repeated completing the square:
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 2 }{ 3 } & 1 & 0 \\
\frac{ 4 }{ 5 } & - \frac{ 6 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 2 }{ 3 } & \frac{ 4 }{ 5 } \\
0 & 1 & - \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 0 \\
0 & 0 & \frac{ 3 }{ 5 } \\
\end{array}
\right)
$$
$$ $$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 2 }{ 3 } & 1 & 0 \\
0 & \frac{ 6 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 0 \\
0 & 0 & \frac{ 3 }{ 5 } \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 2 }{ 3 } & 0 \\
0 & 1 & \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = \left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
$$
==============================================
$$ E_{1} = \left(
\begin{array}{rrr}
1 & - \frac{ 2 }{ 3 } & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$ P_{1} = \left(
\begin{array}{rrr}
1 & - \frac{ 2 }{ 3 } & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{1} = \left(
\begin{array}{rrr}
1 & \frac{ 2 }{ 3 } & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{1} = \left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
$$
==============================================
$$ E_{2} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & - \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$ P_{2} = \left(
\begin{array}{rrr}
1 & - \frac{ 2 }{ 3 } & \frac{ 4 }{ 5 } \\
0 & 1 & - \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{2} = \left(
\begin{array}{rrr}
1 & \frac{ 2 }{ 3 } & 0 \\
0 & 1 & \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{2} = \left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 0 \\
0 & 0 & \frac{ 3 }{ 5 } \\
\end{array}
\right)
$$
==============================================
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 2 }{ 3 } & 1 & 0 \\
\frac{ 4 }{ 5 } & - \frac{ 6 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - \frac{ 2 }{ 3 } & \frac{ 4 }{ 5 } \\
0 & 1 & - \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 0 \\
0 & 0 & \frac{ 3 }{ 5 } \\
\end{array}
\right)
$$
$$ $$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 2 }{ 3 } & 1 & 0 \\
0 & \frac{ 6 }{ 5 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 0 & 0 \\
0 & \frac{ 5 }{ 3 } & 0 \\
0 & 0 & \frac{ 3 }{ 5 } \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 2 }{ 3 } & 0 \\
0 & 1 & \frac{ 6 }{ 5 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
3 & 2 & 0 \\
2 & 3 & 2 \\
0 & 2 & 3 \\
\end{array}
\right)
$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$