Translate and rotate the coordinate axes to put the quadric in standard position, $2xy+2xz+2yz-6x-6y-4z=-9$

Question

Referring to the image attached. Is my working correct?

The answer gives the coefficient on the other side of the equation is -1. Please show where went wrong.

Thanks.

Answer 1

Let $A = \begin{pmatrix} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{pmatrix} \, . $ To write the quadric in standard form, we must write it with respect to an orthonormal set of eigenvectors, which amounts to orthogonally diagonalizing $A$. As mentioned in the comments, you need to apply Gram-Schmidt orthogonalization to obtain an orthonormal basis for the eigenspace of the eigenvalue $-1$.

Diagonalizing $A$, we find that $P^{-1} A P = D$ where $$ D = \left(\begin{array}{rrr} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right) \qquad P = \left(\begin{array}{rrr} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & -1 \end{array}\right) \, . $$ Let $w_i$ be the $i^\text{th}$ column of $P$. Note that the last two columns $w_2, w_3$ are not orthogonal. Applying Gram-Schmidt, we set $v_2 = w_2$ and compute $$ v_3 = w_3 - \operatorname{proj}{v_2}(w_3) = w_3 - \frac{w_3 \cdot v_2}{v_2 \cdot v_2} v_2 = w_3 - \frac{1}{2} v_2 = \begin{pmatrix} -1/2\\ 1\\ -1/2 \end{pmatrix} \, . $$ Normalizing the vectors, we obtain the orthogonal matrix $$ Q = \left(\begin{array}{rrr} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & 0 & \frac{\sqrt{2}}{\sqrt{3}}\\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} \end{array}\right) \, . $$ To deal with the degree $1$ terms, we compute $$ \left(-6,\,-6,\,-4\right) \left(\begin{array}{rrr} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & 0 & \frac{\sqrt{2}}{\sqrt{3}}\\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} \end{array}\right) = \left(-\frac{16}{\sqrt{3}}, -\sqrt{2},\,-\frac{\sqrt{2}}{\sqrt{3}}\right) \, . $$ Thus the transformed quadric is $0 = 2 x^2 - y^2 - z^2 - \frac{16}{\sqrt{3}} x - \sqrt{2}y -\frac{\sqrt{2}}{\sqrt{3}} z + 9$ and completing the square yields \begin{align*} 0 &= 2 x^2 - y^2 - z^2 - \frac{16}{\sqrt{3}} x - \sqrt{2}y -\frac{\sqrt{2}}{\sqrt{3}} z + 9\\ &= 2\left(\left(x - \frac{4}{\sqrt{3}} \right)^2 - \frac{16}{3} \right) - \left(\left(y + \frac{\sqrt{2}}{2} \right)^2 - \frac{1}{2} \right) - \left(\left(z + \frac{1}{\sqrt{6}} \right)^2 - \frac{1}{6} \right) + 9\\ &= 2 u^2 - v^2 - w^2 - \frac{32}{3} + \frac{1}{2} + \frac{1}{6} + 9 = 2 u^2 - v^2 - w^2 - 1 \, . \end{align*}