How to reduce quadratic form using orthogonal transformation

Question

I wanted to reduce following quadratic form $f(x,y,z)=3x^2+3y^2+3z^2-xy-yz$

I found its symmetric metric but its eigenvalues too weird to start to for basis that convert to Eigen basis.

I am new to this topic I had just done diagonalization theory.

Please If some one suggested me how to tackle such problem I would be thankful

Answer 1

One can do this without using orthogonal matrices,

$$ H = \left( \begin{array}{rrr} 6 & - 1 & 0 \\ - 1 & 6 & - 1 \\ 0 & - 1 & 6 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 6 & - 1 & 0 \\ - 1 & 6 & - 1 \\ 0 & - 1 & 6 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 6 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 6 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 6 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 6 & 0 & 0 \\ 0 & \frac{ 35 }{ 6 } & - 1 \\ 0 & - 1 & 6 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 6 }{ 35 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 6 } & \frac{ 1 }{ 35 } \\ 0 & 1 & \frac{ 6 }{ 35 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 6 } & 0 \\ 0 & 1 & - \frac{ 6 }{ 35 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 6 & 0 & 0 \\ 0 & \frac{ 35 }{ 6 } & 0 \\ 0 & 0 & \frac{ 204 }{ 35 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 6 } & 1 & 0 \\ \frac{ 1 }{ 35 } & \frac{ 6 }{ 35 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 6 & - 1 & 0 \\ - 1 & 6 & - 1 \\ 0 & - 1 & 6 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 6 } & \frac{ 1 }{ 35 } \\ 0 & 1 & \frac{ 6 }{ 35 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 6 & 0 & 0 \\ 0 & \frac{ 35 }{ 6 } & 0 \\ 0 & 0 & \frac{ 204 }{ 35 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 6 } & 1 & 0 \\ 0 & - \frac{ 6 }{ 35 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 6 & 0 & 0 \\ 0 & \frac{ 35 }{ 6 } & 0 \\ 0 & 0 & \frac{ 204 }{ 35 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 6 } & 0 \\ 0 & 1 & - \frac{ 6 }{ 35 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 6 & - 1 & 0 \\ - 1 & 6 & - 1 \\ 0 & - 1 & 6 \\ \end{array} \right) $$