Expressing a projective conic in standard form

Question

I was given the task to find a projective transformation $\phi: \mathbb{P}_\mathbb{C} \rightarrow \mathbb{P}_\mathbb{C}$ that transforms the following conic:

$$2\,{x}^{2}+2\,xy-3\,xz+4\,yz+{z}^{2}=0$$

into the form:

$$X^2 + Y^2 + Z^2 = 0$$

Then my attempt to do so was to first find the matrix of the symmetric bilinear form which is given by:

$$ \left[ \begin {array}{ccc} 2&1&-3/2\\ 1&0&2 \\ -3/2&2&1\end {array} \right] $$

Then I wanted to orthogonally diagonalise this matrix to obtain an expression that doesn't have any cross terms in it. However the eigenvectors of this matrix are not "nice" (I've checked with Maple and they involve a lot of square roots).

In the solution given to me, the matrix that expresses this conic in standard form was given by the following matrix:

$$ \left[ \begin {array}{ccc} 1/2\,\sqrt {2}&1/10\,\sqrt {10}&-1/3\, \sqrt {3}\\ 0&1/10\,\sqrt {10}&-i/6\sqrt {3} \\ 0&1/5\,\sqrt {10}&i/3\sqrt {3}\end {array} \right] $$

I can see that this matrix is orthogonal, but how does it simplify the expression of the conic above? And how do I find it from scratch? And how do this in general?

Answer 1

I found some notes online, the task is to begin with symmetric $H$ and solve $P^T HP = D$ diagonal, which can be done over the reals. Then forcing the identity over the complexes requires multiplying by an additional diagonal complex matrix on both sides.

Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ - 2 & \frac{ 11 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 2 & - 3 \\ 2 & 0 & 4 \\ - 3 & 4 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - 2 \\ 0 & 1 & \frac{ 11 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 30 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ - \frac{ 3 }{ 4 } & - \frac{ 11 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 30 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & - \frac{ 3 }{ 4 } \\ 0 & 1 & - \frac{ 11 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 2 & - 3 \\ 2 & 0 & 4 \\ - 3 & 4 & 2 \\ \end{array} \right) $$

Alright, given this $P$ and the resulting $D,$ we multiply on the right and left by $$ P_1 = \left( \begin{array}{rrr} \frac{1}{2} & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & \frac{1}{\sqrt {30}} \\ \end{array} \right) $$

to reach $I.$ I began with double your symmetric matrix, so you may need to multiply throughout by $\sqrt 2$ to take your actual matrix to $I.$ Note how completely the answer changes based on choices made; mine looks nothing like your given answer.

Answer 2

So you want to diagonalize $$A=\begin{bmatrix} 2&1&-3/2\\ 1&0&2 \\ -3/2&2&1\end{bmatrix}$$

First find the eigenvalues

$$ {\rm det}(A-w \mathbf{1}) = 0 $$

which are $$\mathbf{w} = \pmatrix{w_1 \\ w_2 \\ w_3} = \pmatrix{ -2.2599587007 \\ 3.1586587920 \\ 2.1012999087} $$ in arbitrary order.

Now form a matrix $X$ from the eigenvectors as columns. There are many ways to get there, but I will demonstrate a linear algebra method.

Solve the system $$ (A -w \mathbf{1}) \pmatrix{v_1\\v_2\\v_3} = \pmatrix{ \xi \\ \xi \\ \xi } $$ for $v_1$, $v_2$ and $\xi$ and defer the $\xi \rightarrow 0$ for later.

This gives me the eigenvector as a function of $w$

$$ \mathbf{v}(w)= \pmatrix{v_1 \\ v_2 \\ v_3} = \pmatrix{ \left( \frac{2 w^2-3 w-12}{2 w^2-3 w-9} \right) v_3 \\ \left(1+ \frac{ 3(2 w-7)}{2 ( 2 w^2-3 w-9)} \right) v_3 \\ v_3} $$

and we need to normalize the eigenvectos, such that $\mathbf{v}^\top \mathbf{v} = 1$ which leads to $$v_3^2 = \frac{1}{3+ \left( \frac{3(16 w^3-84 w^2-48w+483)}{4(2 w^2-3w-9)^2}\right)}$$

Now that we can define the eigenvectors $\mathbf{v}$ for each eigenvalue $w$ we construct the following 3×3 matrix

$$ X = \begin{bmatrix} \vdots & \vdots & \vdots \\ \mathbf{v}(w_1) & \mathbf{v}(w_2) & \mathbf{v}(w_3) \\ \vdots & \vdots & \vdots \end{bmatrix} $$

$$ X = \begin{bmatrix} 0.37748124858 & -0.70137614012 & -0.60463163747 \\ -0.70173922139 & 0.20936797859 & -0.68097512046 \\ 0.60421020524 & 0.68134907323 & -0.41315066050 \end{bmatrix} $$

And now for the determination of the diagonal matrix $D$ which factors $A$ as follows $$A = X D X^{-1}$$ or $$D = X^{-1} A X$$

$$D = \begin{bmatrix} -2.2599587007 & & \\ & 3.1586587920 & \\ & & 2.1012999087 \end{bmatrix}$$

This means a transformation for the solution vector $p = \pmatrix{x \\ y \\ z}$ into $$p = X u $$ or $$ u = X^{-1} p$$

The conic

$$ p^\top A p = \mathbf{0} $$

becomes

$$ u ^\top D u = \mathbf{0} $$

since $X^\top = X^{-1}$ and $p^\top A p = (X u)^\top X D X^{-1} (X u) = u^\top (X^\top X ) A ( X^{-1} X) u = u^\top D u$.

Answer 3

We can compute the Jordan decomposition: $$ \begin{align} M &=\begin{bmatrix} 2&1&-\frac32\\ 1&0&2\\ -\frac32&2&1 \end{bmatrix}\tag{1a}\\[9pt] &=PDP^{-1}\tag{1b} \end{align} $$ where $$ D=\begin{bmatrix} -2.25995870076 & 0 & 0 \\ 0 & 2.10129990875 & 0 \\ 0 & 0 & 3.15865879201 \end{bmatrix}\tag2 $$ and $$ P=\begin{bmatrix} 0.377481248580 & 0.604631637473 & -0.701376140125 \\ -0.701739221400 & 0.680975120463 & 0.209367978594 \\ 0.604210205243 & 0.413150660507 & 0.681349073238 \end{bmatrix}\tag3 $$ We compute the Jordan decomposition by computing the eigenvalues and eigenvectors of $M$; note that $\det(M-I\lambda)=-\frac14\left(4\lambda^3-12\lambda^2-21\lambda+60\right)$. Since $MP=PD$, the columns of $P$ are eigenvectors of $M$ (so they are orthogonal) and the eigenvalues are the diagonal elements of $D$. Normalize the columns of $P$ (that is, pick unit eigenvectors) so that $P^TP=I$.

Set $$ \begin{align} \begin{bmatrix} u\\v\\w \end{bmatrix} &=P^T \begin{bmatrix} x\\y\\z \end{bmatrix}\tag{4a}\\ &=\begin{bmatrix} 0.377481248580 & -0.701739221400 & 0.604210205243 \\ 0.604631637473 & 0.680975120463 & 0.413150660507 \\ -0.701376140125 & 0.209367978594 & 0.681349073238 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix}\tag{4b} \end{align} $$ Then $$ \begin{align} \begin{bmatrix} u\\v\\w \end{bmatrix}^T D \begin{bmatrix} u\\v\\w \end{bmatrix} &= \begin{bmatrix} x\\y\\z \end{bmatrix}^T PDP^T \begin{bmatrix} x\\y\\z \end{bmatrix}\tag{5a}\\ &= \begin{bmatrix} x\\y\\z \end{bmatrix}^T M \begin{bmatrix} x\\y\\z \end{bmatrix}\tag{5b} \end{align} $$ That is, $$ \begin{align} &-2.25995870076u^2+2.10129990875v^2+3.15865879201w^2\\ &=2x^2+2xy-3xz+4yz+z^2\tag6 \end{align} $$