Principal Axis Transformation

Quadratic Forms with Symmetric Matrices

A quadratic equation with a $cross term$ does not state its principal directions directly. Its quadratic form collects the coefficients in a symmetric matrix, and the eigenvectors of that matrix give the principal directions. In two dimensions, let

A = \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix}

Because $A$ is real and symmetric, it has real eigenvalues $\lambda_1,\lambda_2$ and an orthonormal eigenbasis $v_1,v_2$ . Put those eigenvectors into $S=(v_1\quad v_2)$ . Then $S^TS=I$ and $S^TAS=\operatorname{diag}(\lambda_1,\lambda_2)$ .

For the equation $x^TAx+p^Tx+f=0$ , use the rotated coordinates $\xi=S^Tx$ and the rotated linear coefficient $(\delta,\epsilon)^T=S^Tp$ . The cross term disappears:

\lambda_1 \xi_1^2 + \lambda_2 \xi_2^2 + \delta \xi_1 + \epsilon \xi_2 + f = 0

Completing the Square

For both variables with $\lambda_1, \lambda_2 \neq 0$ , the completing the square process is performed separately:

Name the completed-square brackets to keep both $added and subtracted terms$ visible:

B_1=\xi_1^2+2\frac\delta{2\lambda_1}\xi_1 +\frac{\delta^2}{4\lambda_1^2}

Collecting the completed squares gives

\begin{aligned} 0 &= \lambda_1 \left( \xi_1 + \frac{\delta}{2\lambda_1} \right)^2 \\ &\quad{}+ \lambda_2 \left( \xi_2 + \frac{\epsilon}{2\lambda_2} \right)^2 \\ &\quad{}+ \left( f - \frac{\delta^2}{4\lambda_1} - \frac{\epsilon^2}{4\lambda_2} \right) \end{aligned}

Define the center in rotated coordinates and the translated constant by

(m_1,m_2)=\left(-\frac{\delta}{2\lambda_1},-\frac{\epsilon}{2\lambda_2}\right)

Then

\begin{aligned} &\lambda_1 (\xi_1 - m_1)^2 + \lambda_2 (\xi_2 - m_2)^2 - \gamma \\ &= 0 \end{aligned}

The signs of $\lambda_1$ , $\lambda_2$ , and $\gamma$ determine whether the real solution set is an ellipse, a hyperbola, a parabola, a degenerate conic, or empty. The next two cases assume $\gamma>0$ .

Curve Classification

With $positive translated constant$ $\gamma$ , two nonzero eigenvalues of the same positive sign give an ellipse, while opposite signs give a hyperbola. A zero eigenvalue requires checking the remaining linear term: it yields a parabola only when that term is nonzero.

Two Positive Eigenvalues

If $\lambda_1 > 0$ and $\lambda_2 > 0$ , then the conic section formed is an $ellipse$ :

\frac{(\xi_1 - m_1)^2}{r_1^2} + \frac{(\xi_2 - m_2)^2}{r_2^2} = 1

With semi-axis lengths $r_1 = \sqrt{\frac{\gamma}{\lambda_1}}$ in the direction of $v_1$ and $r_2 = \sqrt{\frac{\gamma}{\lambda_2}}$ in the direction of $v_2$ .

Ellipse Visualization in Coordinates

\xi_1, \xi_2

Ellipse curve with both positive eigenvalues and principal axes aligned with eigenvector directions.

$x$: $[-4, 4]$
$y$: $[-4, 4]$

Eigenvalues with Opposite Signs

When $\lambda_1 > 0$ and $\lambda_2 < 0$ , the conic section formed is a $hyperbola$ :

\frac{(\xi_1 - m_1)^2}{r_1^2} - \frac{(\xi_2 - m_2)^2}{r_2^2} = 1

With semi-axis lengths $r_1 = \sqrt{\frac{\gamma}{\lambda_1}}$ in the direction of $v_1$ and $r_2 = \sqrt{\frac{\gamma}{-\lambda_2}}$ in the direction of $v_2$ .

Hyperbola Visualization in Coordinates

\xi_1, \xi_2

Hyperbola curve with principal axes aligned with eigenvector directions and center transformation.

$x$: $[-4, 4]$
$y$: $[-4.24, 4.24]$

One Zero Eigenvalue

A special case occurs when $\lambda_1\neq0$ , $\lambda_2=0$ , and $\epsilon\neq0$ . Completing the square gives

Keep the same completed-square brackets visible, because the added and subtracted terms must cancel separately:

B_1=\xi_1^2+2\frac\delta{2\lambda_1}\xi_1 +\frac{\delta^2}{4\lambda_1^2}

Here the zero eigenvalue means we use only the first completed square, so define:

m_1=-\frac{\delta}{2\lambda_1},\qquad \gamma=\frac{\delta^2}{4\lambda_1}-f

Solving for $\xi_2$ shows the $parabola directly$ :

\xi_2 = -\frac{\lambda_1}{\epsilon} (\xi_1 - m_1)^2 + \frac{\gamma}{\epsilon}

Parabola Visualization in Coordinates

\xi_1, \xi_2

Parabola curve with one zero eigenvalue and coordinate axis transformation aligned with eigenvectors.

$x$: $[-3.5, 3.5]$
$y$: $[-1, 4]$

Rotating and Translating a Two Dimensional Conic

Consider

5x_1^2+4x_1x_2+2x_2^2-6x_1-9=0

The eigenpairs are

\begin{aligned} &\lambda_1=6, \\ &v_1=\frac1{\sqrt5}\begin{pmatrix}2\\1\end{pmatrix}, \\ &\lambda_2=1, \\ &v_2=\frac1{\sqrt5}\begin{pmatrix}-1\\2\end{pmatrix} \end{aligned}

The $center in the original coordinates$ is $h=(1,-1)^T$ because $2Ah+p=0$ . In rotated coordinates, $m=S^Th=(1/\sqrt5,-3/\sqrt5)^T$ . With $\eta=\xi-m$ , the equation becomes

6\eta_1^2+\eta_2^2=12

It is an ellipse centered at $(1,-1)$ . Its semiaxes have lengths $\sqrt2$ along $v_1$ and $2\sqrt3$ along $v_2$ . Substituting the center into the original equation gives $-12$ , which confirms the translated right-hand side.

Quadratic Surfaces and Transformation

For a symmetric matrix $A \in \mathbb{R}^{n \times n}$ , vector $b \in \mathbb{R}^n$ , and scalar $c \in \mathbb{R}$ , the $quadratic surface$ $Q$ is defined as the solution set of the general quadratic equation:

x^T A x + b^T x + c = 0

Writing the same equation entry by entry gives:

\sum_{j=1}^n\sum_{k=1}^n x_ja_{jk}x_k +\sum_{j=1}^n b_jx_j+c=0

If $A \in \mathbb{R}^{n \times n}$ is symmetric and $v_1, \ldots, v_n \in \mathbb{R}^n$ is an orthonormal basis of eigenvectors with $A \cdot v_i = \lambda_i \cdot v_i$ , then the orthonormal matrix $S = (v_1 \quad \ldots \quad v_n)$ enables diagonalization $A \cdot S = S \cdot \Lambda$ or $\Lambda = S^{-1} \cdot A \cdot S = S^T \cdot A \cdot S$ .

In the rotated coordinates $\xi=S^Tx$ , write $\mu=S^Tb$ . The coordinates $\xi\in\mathbb R^n$ of points on $Q$ satisfy:

\xi^T S^T A S \xi + b^T S \xi + c = 0

Diagonalizing $A$ gives:

\xi^T\Lambda\xi+\mu^T\xi+c=0

Written entry by entry:

\sum_{j=1}^n\lambda_j\xi_j^2 +\sum_{j=1}^n\mu_j\xi_j+c=0

Collecting the terms and moving $c$ to the right gives:

\sum_{j=1}^n (\lambda_j \xi_j^2 + \mu_j \xi_j) = -c

In the orthonormal basis of eigenvectors, the quadratic form has a diagonal structure. This transformation is called $principal axis transformation$ because the new coordinate axes are $aligned with the directions of the matrix eigenvectors$ .