I have got myself into a pickle with the following question:
Classify the following (ellipse, hyperbola, ellipsoid etc)
$x^2 + y^2 + 2z^2 + 2xz - 2y + 2z + 2 =0$
Now, I have written a symmetric matrix $A = \left(\begin{array}{ccc}1&0&1\\0&1&0\\1&0&2\\\end{array}\right)$ and have calculated eigenvalues $\lambda_1 =1, \lambda_2 =\frac12(3+\sqrt5),\lambda_3 =\frac12(3-\sqrt5).$
Corresponding eigenvectors are $v_1 = (0,1,0)^T, v_2 =(\sqrt5-1,0,2)^T,v_3 =(1+\sqrt5,0,-2)^T$
Then by dividing my their respective lengths we obtain a Change of Basis Matrix $P$ such that $P^TAP$ has the eigenvalues $\lambda_1,\lambda_2,\lambda_3$ in the diagonal entries.
$P = \left(\begin{array}{ccc}0&B&C\\1&0&0\\0&C&-B\\\end{array}\right)$ With $B$ and $C$ are defined as
$B=\sqrt{\frac{1}{10}(5-\sqrt5)}$ and $C=\sqrt{\frac{1}{10}(5+\sqrt5)}$
I have then established that we have the following change of co-ordinates, corresponding to $(x,y,z)^T = P(x_1,y_1,z_1)^T$.
We then obtain that $x_1^2 + \frac12(3+\sqrt5)y_1^2 + \frac12(3-\sqrt5)z_1^2 -2x_1+2Cy_1-2Bz_1+2=0$
Finally, we complete the square on $x_1,y_1,z_1$ to obtain that
$x_2^2 +\frac12(3+\sqrt5)y_2^2 + \frac12(3-\sqrt5)z_2^2 = 1$
Since the coefficients are all positive, this seems to specify an ellipsoid.
However, the problem is that typing in the equation $x^2 + y^2 + 2z^2 + 2xz - 2y + 2z + 2 =0$ (into Wolfram Alpha) doesn't seem to tell me that I have an ellipsoid.
More worryingly, I have noticed that $x^2 + y^2 + 2z^2 + 2xz - 2y + 2z + 2 = (x+z)^2 + (y-1)^2 + (z+1)^2$ and this all seems too convienient for me to have ignored this.
If anyone could please let me know if I am on the right track or if not, then how I could manage to find the right track again, I would be very grateful.
Thanks guys