Complete the square on cross terms?

Question

I am trying to complete the square on $xy - xz + yz = 1$.

I have tried plugging in $u = x + y$. But I still end up with unsolvable cross terms.

Answer 1

All you can do is $y(x+z)=1$ then divide by one or the other if it suits you. There is no square to complete.

Answer 2

Summary: the surface $xy-zx+yz=1$ is a rotated version of the surface of revolution $x^2 + y^2 = 1 + 2 z^2,$ which is a hyperboloid of one sheet.

The first matrix identity below, $Q^T DQ=H,$ says that $$ \frac{1}{4}(x+y)^2 - \frac{1}{4}(-x+y-2z)^2 + z^2 = xy - zx + yz \; . $$ This displays the three variables $$ u=x+y, \; \; v = -x + y - 2 z, \; \; w = z \; . $$ Sylvester's Law of Inertia says that there are two positive and one negative eigenvalue of the Hessian matrix of $xy-zx + yz.$ Setting it equal to $1$ gives a hyperboloid of one sheet.

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} \frac{ 1 }{ 2 } & - 1 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ 0 & - 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ - 1 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & 1 & - 1 \\ 1 & 0 & 1 \\ - 1 & 1 & 0 \\ \end{array} \right) $$

Also, we can use the fact that the eigenvalues of the original matrix are integers to find an orthogonal matrix that gives us

$$\left( \begin{array}{rrr} \frac{ 1 }{ \sqrt 2 } & - \frac{ 1 }{ \sqrt 6 } & \frac{ 1 }{ \sqrt 3 } \\ \frac{ 1 }{ \sqrt 2 } & \frac{ 1 }{ \sqrt 6 } & - \frac{ 1 }{ \sqrt 3 } \\ 0 & \frac{ 2 }{ \sqrt 6 } & \frac{ 1 }{ \sqrt 3 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \\ \end{array} \right) \left( \begin{array}{rrr} \frac{ 1 }{\sqrt 2 } & \frac{ 1 }{ \sqrt2 } & 0 \\ - \frac{ 1 }{\sqrt 6 } & \frac{ 1 }{\sqrt 6 } & \frac{ 2 }{\sqrt 6 } \\ \frac{ 1 }{\sqrt 3 } & -\frac{ 1 }{\sqrt 3 } & \frac{ 1 }{\sqrt 3 } \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & 1 & - 1 \\ 1 & 0 & 1 \\ - 1 & 1 & 0 \\ \end{array} \right) $$ which shows that the hyperboloid is a surface of revolution

Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
$$ H = \left( \begin{array}{rrr} 0 & 1 & - 1 \\ 1 & 0 & 1 \\ - 1 & 1 & 0 \\ \end{array} \right) $$ $$ D_0 = H $$

$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

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

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

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

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - 1 \\ 1 & \frac{ 1 }{ 2 } & 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ - 1 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 2 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 1 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ - 1 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 1 & - 1 \\ 1 & 0 & 1 \\ - 1 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - 1 \\ 1 & \frac{ 1 }{ 2 } & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 2 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} \frac{ 1 }{ 2 } & - 1 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ 0 & - 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ - 1 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & 1 & - 1 \\ 1 & 0 & 1 \\ - 1 & 1 & 0 \\ \end{array} \right) $$