How to reduce a quadric into canonical form without rigorous calculations?

Question

Consider the quadric in $(x,y,z)$ given by $x^2+y^2+z^2-2xy-2yz+2zx+x-4y+z+1=0$.I am asked to reduce it into canonical form and describe the nature.Here it is a parabolic cylinder.But the method taught to us by the college prof. is based on calculation of eigenvalues and unit eigenvectors,then applying a rotation.But this becomes very long,I am looking for a smarter approach to this problems which is based less on calculation and more on linear algebra concepts.I thought first that I can use congruence operation for this but I found that normal form under congruence is not the same as the canonical form of a quadric.

Answer 1

The first six terms can clearly be seen as the expansion of $(x-y+z)^2$, now apply completing square method. $$(x-y+z)^2+x-4y+z+1=0$$ $$\implies (x-y+z)^2+(x-y+z)-3y+1=0$$ $$\implies \Big(x-y+z+\frac{1}{2}\Big)^2=3\Big(y-\frac{1}{4}\Big)$$ which is the required canonical form.

Answer 2

Even if this is (now) an old question, for completeness I would like to give an answer.

As pointed out in the comments, rewriting the quadric as $$\Big(x-y+z+\frac{1}{2}\Big)^2=3\Big(y-\frac{1}{4}\Big)$$ makes clear that we are dealing with a parabolic cylinder, but we want a (metric) canonical form, we have to do something more.

First, let's try to perform all the calculations. Let $B$ be the "complete" matrix of the quadric, i.e. $$ B=\left( \begin{matrix}1 & -1 & 1 & \frac{1}{2}\\ -1 & 1 & -1 & -2\\ 1 & -1 & 1 & \frac{1}{2}\\ \frac{1}{2} & -2 & \frac{1}{2} & 1\end{matrix} \right) $$ and $A$ the of the "quadratic part", i.e. $$ A=\left( \begin{matrix}1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{matrix} \right) $$ Since the eigenvalues of $A$ are $3$, $0$ and $0$ (I mean, $0$ has algebraic multiplicity $2$) and an orthonormal basis of $\mathbb{R}^3$ made by corresponding eigenvectors is $\{\frac{1}{\sqrt{3}}(1,-1,1),\frac{1}{\sqrt{2}}(1,1,0),\frac{1}{\sqrt{6}}(-1,1,2)\}$, then the matrix $$ R=\left( \begin{matrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}} \end{matrix} \right) $$ represents a rotation that trasform that avoid the "rectangular" term from the quadric. Indeed, if we perform the substitution $$ \left( \begin{matrix} x \\ y \\ z \end{matrix} \right)=R\left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) $$ we get the equation (of course, it is easy if one first notices that $x-y+z=\sqrt{3}X$) $$ 3X^2+2\sqrt{3}X-\sqrt{6}Z+1=0. $$ Now, completing the square, we get $$ 3(X+\frac{\sqrt{3}}{3})^2-\sqrt{6}(Z-\frac{2}{\sqrt{6}})=0. $$ and then, with the obvious translation, finally $$ 3X^2-\sqrt{6}Z=0. $$

Now, let's see if we can avoid some annoying calculations. First of all, it's is well known that the nature of a quadric depends on the inertia of $B$ and $A$, i.e., from the number of positive, negative and zero eigenvalues of $B$ and of $A$. Inertia can be easily calculeted from Descartes' rule of signs: in this case we have for $B$ the inertia $(2,1)$ ($2$ positive eigenvalues, $1$ negative and $4-2-1=0$ equal to $0$) and for $A$, $(1,0)$, so a parabolic cylinder. Such a quadric has a canonical form like $$ \Gamma:\ \lambda X^2+2pY=0 $$ with $\lambda$ and $p$ different from $0$ and $\lambda$ positive (the $2$ in $2p$ just serves to avoid fractions). Indeed, if $B'$ and $A'$ are the matrix of $F$, i.e. $$ B'=\left( \begin{matrix}\lambda & 0 & 0 & 0\\ 0 & 0 & 0 & p\\ 0 & 0 & 0 & 0\\ 0 & p & 0 & 1\end{matrix} \right) $$ and $$ A'=\left( \begin{matrix}\lambda & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) $$ then the inertia of $B'$ is $(2,1)$ and that of $A$ is $(1,0)$, as one expects.

We also know that there are four invariants under translation and rotation of the coordinate system, namely, if a quadric has complete matrix $$ \mathcal{B}=\left( \begin{matrix} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{11} & a_{22} & a_{23} & a_{24} \\ a_{13} & a_{23} & a_{33} & a_{34} \\ a_{14} & a_{24} & a_{34} & a_{44} \end{matrix} \right) $$ then the quantities $\det(B)$, $\det(A)$, $\mathrm{tr}(A)$ and $$ I_2=\det\left( \begin{matrix} a_{11} & a_{12}\\ a_{11} & a_{22} \end{matrix} \right)+\det\left( \begin{matrix} a_{22} & a_{23}\\ a_{23} & a_{33} \end{matrix} \right)+\det\left( \begin{matrix} a_{33} & a_{13}\\ a_{13} & a_{11} \end{matrix} \right) $$ do not change under translations and rotations. From this, and in particular from $\mathrm{tr}(A)=\mathrm{tr}(A')$, one can easily obtain $\lambda=3$, but there is no way to get a value for $p$.

Anyway, there are also two other quantities, which, unluckily, are invariant only under rotations, but not under traslations, namely the semi-invariants $$ \Delta_B=B_{11}+B_{22}+B_{33} $$ (where $B_{ij}$ is the algebraic complement of the element $a_{ij}$ in $B$) and $$ \Delta'_B=\det\left( \begin{matrix} a_{11} & a_{14}\\ a_{14} & a_{44} \end{matrix} \right)+\det\left( \begin{matrix} a_{22} & a_{14}\\ a_{14} & a_{44} \end{matrix} \right)+\det\left( \begin{matrix} a_{33} & a_{34}\\ a_{34} & a_{44} \end{matrix} \right). $$

Now, if we set $\Delta_B=\Delta_{B'}$ and $\Delta'_B=\Delta'_{B'}$, we get $$ \left\{ \begin{array}{l} 3-p^2=-\frac{3}{2}\\ 3p^2=\frac{9}{2} \end{array} \right. $$ which is impossible, and so clearly wrong. Why?

As said, $\Delta_B$ and $\Delta'_B$ are invariants only under rotation, but to bring the quadric in canonical form, we also need a traslation. But we know that after the rotation, the quadric becomes something like $3X^2+2aX+2pY=0$, as in general we cannot avoid a linear term for $X$ (again, the $2a$ is just for convenience). Let $B''$ be the matrix for this quadric, so $$ B''=\left( \begin{matrix}3 & 0 & 0 & a\\ 0 & 0 & 0 & p\\ 0 & 0 & 0 & 0\\ a & p & 0 & 1\end{matrix} \right) $$ then from $\Delta_B=\Delta_{B''}$ and $\Delta'_B=\Delta'_{B''}$, we get the system $$ \left\{ \begin{array}{l} (3-a^2)-p^2=-\frac{3}{2}\\ 3p^2=\frac{9}{2} \end{array} \right. $$ from which we have $p=\pm\frac{\sqrt{3}}{2}$: since we do not need the value of $a$, the only useful equation is $\Delta'_B=\Delta'_{B''}$.

Lastly, there is also a third possibility. Let $\lambda^4-D_1\lambda^3+D_2\lambda^2-D_3\lambda+D_4$ the characteristic polynomial of $B$ and $\lambda^3-d_1\lambda^2+d_2\lambda-d_3$ the characteristic polynomial of $A$, important: the two polynomial bust be written in order to be monic.

We have the following: if our quadric is a elliptic (possibly imaginary) or hyperbolic cylinder, then its canonical form is $$ \lambda_1X^2+\lambda_2Y^2+\frac{D_3}{d_2}=0 $$
(where $\lambda_1$ and $\lambda_2$ are the non zero eigenvalues of $A$) while if it is a parabolic cylinder, then a canonical form is $$ \lambda_1X^2-2Y\sqrt{\frac{-D_3}{d_1}}=0 $$ With the same rationale, in the case of a pair of parallel straight lines, a canonical form is $$ \lambda_1X^2+\frac{D_2}{d_1}=0 $$ (again, here and above, $\lambda_1$ is the non zero eigenvalue of $A$).