In the three variables $x$, $y$ and $z$, the most general quadratic form is $$Q(x,y,z) = ax^2 + by^2 + cz^2 + dxy + exz + fyz,$$ where in both cases $a,b,c,d,e,f$ are all constants. It is possible to express quadratic forms in $n$ variables as a matrix product $\v^TA\v$, where $\v$ is a vector with the $n$ variables as entries and $A$ is a symmetric matrix.

The two variable quadratic form above can be expressed as $$Q(x,y) = \mat{cc}{x & y}\mat{cc}{a & c/2 \\ c/2 & b}\mat{c}{x\\ y}.$$

The three variable quadratic form given above can be written as $$Q(x,y,z) = \mat{ccc}{x & y & z}\mat{ccc}{a & d/2 & e/2\\ d/2 & b & f/2\\ e/2 & f/2 & c}\mat{c}{x\\ y\\ z}.$$ As an exercise, trying verifying this by expanding out both expressions. Observe that in both cases the diagonal entries of the matrix are the coefficients of the square terms and the off-diagonal entries in the matrix are the coefficients of the cross-terms.

Note that the expression $2x^2+6xy-7y^2$ can be written as $$\underbrace{ \left( \begin{array}{rr} x & y \end{array} \right) }_{{\large \v^T}} \underbrace{ \left( \begin{array}{rr} 2 & 3 \\ 3 & -7 \end{array} \right) }_{{\large A}} \underbrace{ \left( \begin{array}{r} x \\ y \end{array} \right) }_{{\large \v}}$$ The matrix $A$ is symmetric. Then it is a unique symetric matrix representation.

First, set $Q=-3x^2-2y^2-3z^2+2xy+2yz.$ Here we want to find a change of variables $$\left( \begin{array}{r} x \\ y \\ z \end{array} \right) = P \left( \begin{array}{r} u \\ v \\ w \end{array} \right)$$ where $P$ is an orthogonal matrix, such that $$Q = \lambda_1u^2 + \lambda_2 v^2 + \lambda_3 w^2$$ in terms of the new variables $u,v,w.$

So we need to find the orthogonal matrix $P$ that orthogonally diagonalises $A$. That is $$P^TAP = D.$$

From Example 18.2.4 we know that $$P = \left( \begin{array}{rrr} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} & \dfrac{1}{\sqrt{3}}\\ 0 & \dfrac{2}{\sqrt{6}} & -\dfrac{1}{\sqrt{3}}\\ -\dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} & \dfrac{1}{\sqrt{3}} \end{array} \right)\;\; \text{ & } \;\; D = \left(\begin{array}{rrr}-3&0&0\\0&-1&0\\0&0&-4\end{array}\right)$$

Then $\ds \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = P \left( \begin{array}{r} u \\ v \\ w \end{array} \right)$ $=\ds \left( \begin{array}{r} \dfrac{u}{\sqrt{2}} + \dfrac{v}{\sqrt{6}} + \dfrac{w}{\sqrt{3}}\\ \dfrac{2v}{\sqrt{6}} -\dfrac{w}{\sqrt{3}}\\ -\dfrac{u}{\sqrt{2}} + \dfrac{v}{\sqrt{6}} + \dfrac{w}{\sqrt{3}} \end{array} \right)\;$

A non-degenerate conic section is in standard position relative to the coordinate axes if its equation can be expressed in one of the following forms:

• $\ds{\frac{x^2}{k^2}+\frac{y^2}{l^2}=1; \;k,l>0,}$
• $\ds{\frac{x^2}{k^2}-\frac{y^2}{l^2}=1}$ or $\ds{\frac{y^2}{l^2}-\frac{x^2}{k^2}=1; \;k,l>0,}$
• $x^2=ky$ or $y^2=kx; \;k\ne0.$

The key observation here is that conic sections in standard form have no cross-terms. Given a quadratic equation with cross-terms in the associated quadratic form, we can change variables to remove the cross-terms by orthogonal diagonalisation.

Due to the defining property of rotation matrices, an orthogonal matrix $P$ always corresponds to a rotation, provided $\det(P)=1$ (not $-1$).

Hence, we have the following:

Another important observation is that there is never an occurance of $x^2$ and $x$ in the standard form (or $y^2$ and $y$). As a general rule, given a quadratic equation (even after changing variables from orthogonal diagonalisation), if we have terms such as $x^2$ and $x$ (or similar terms involving new variables) we can complete the square to be left with only a square term. We have the following:

In summary, to identify a quadratic equation as a conic section, we follow these steps:

In summary, to identify a quadratic equation as a conic section, we follow these steps:

Step 2. Find eigenvalues and eigenvectors of $A$.

Step 2. Find eigenvalues and eigenvectors of $A$.

Step 2. Now we form the matrix $P= \left(\hat{\v}_1 ~|~ \hat{\v}_2 \right),$ i.e., $$P = \dfrac{1}{\sqrt{2}} \left( \begin{array}{r} 1 & 1 \\ 1 & -1 \end{array} \right).$$

But note that $\text{det}(P) = -1.$ This is NOT a rotation! 😕 No problem! We just swap the columns to obtain $$P = \left(\hat{\v}_2 ~|~ \hat{\v}_1\right) = \dfrac{1}{\sqrt{2}} \left( \begin{array}{rr} 1 & 1 \\ -1 & 1 \end{array} \right)$$ so $\text{det}(P) = 1$ and now $P$ is a rotation! 😎

Step 2. Now we form the matrix $P= \left(\hat{\v}_1 ~|~ \hat{\v}_2 \right),$ i.e., $P = \dfrac{1}{\sqrt{2}} \left( \begin{array}{r} 1 & 1 \\ 1 & -1 \end{array} \right).$

But note that $\text{det}(P) = -1.$ This is NOT a rotation! 😕 No problem! We just swap the columns to obtain $$P = \left(\hat{\v}_2 ~|~ \hat{\v}_1\right) = \dfrac{1}{\sqrt{2}} \left( \begin{array}{rr} 1 & 1 \\ -1 & 1 \end{array} \right)$$ so $\text{det}(P) = 1$ and now $P$ is a rotation! 😎

Step 3. We now write the new variable

Step 4. Consider the expression $\mathbf x^T A \mathbf x + K \mathbf x + 4 = 0.$

Step 5. In the expression $\sqrt{2}u + 2v^2 - 4 \sqrt{2} v + 4 = 0$, we complete the square in $v$. That is $$\sqrt{2} u + 2v^2 - 4 \sqrt{2} v \, \underbrace{\,+ \,\;2\left(\sqrt{2}\right)^2 - 2 \left(\sqrt{2}\right)^2}_{{\large \text{adding }0}} + 4 =0.$$ $$\Ra\; \sqrt{2} u + \underbrace{2\left(v^2 - 2 \sqrt{2} v + \left(\sqrt{2}\right)^2\right)}_{{\large \text{factorise }2}} - 2 \left(\sqrt{2}\right)^2 + 4 =0.$$ $$\Ra\; \sqrt{2} u + 2\left(v - \sqrt{2}\right)^2 =0.$$

Step 6. Note that $\sqrt{2} u + 2\left(v - \sqrt{2}\right)^2 =0$ is a non-degenerate conic.

Set $\;s=u\;$ and $\;t = v-\sqrt{2}$. Thus $$\sqrt{2}s + 2t^2 = 0.$$