Real Axis Transformation

Quadratic Forms with Symmetric Matrices

When you encounter a general quadratic equation, the best way to understand it is by looking at its structure in matrix form. Imagine you have a symmetric matrix $A$ with elements:

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

This matrix has eigenvalues $\lambda_1, \lambda_2 \in \mathbb{R}$ and corresponding orthonormal eigenvectors $v_1, v_2 \in \mathbb{R}^2$ . Something interesting happens when we use coordinate transformation through the matrix $S = (v_1 \quad v_2) \in \mathbb{R}^{2 \times 2}$ .

For coordinate transformation, we use $\begin{pmatrix} \delta \\ \epsilon \end{pmatrix} = S^T \begin{pmatrix} d \\ e \end{pmatrix}$ . In the new coordinates $\xi = S^T x$ , the equation becomes:

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

Completing the Square Process

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

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

The result of this process gives a simpler form:

\begin{aligned} &= \lambda_1 \left( \xi_1 + \frac{\delta}{2\lambda_1} \right)^2 + \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}

By determining the center point $(m_1, m_2) = \left( -\frac{\delta}{2\lambda_1}, -\frac{\epsilon}{2\lambda_2} \right)$ and the constant $\gamma = \frac{\delta^2}{4\lambda_1} + \frac{\epsilon^2}{4\lambda_2} - f$ , we obtain:

\lambda_1 (\xi_1 - m_1)^2 + \lambda_2 (\xi_2 - m_2)^2 - \gamma = 0

For $\gamma > 0$ , various curve forms can emerge depending on the signs of the eigenvalues.

Curve Classification

Both Eigenvalues Positive

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.

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.

One Eigenvalue Zero

A special condition occurs when $\lambda_1 \neq 0$ and $\lambda_2 = 0$ . Completing the square gives:

0 = \lambda_1 \left( \xi_1^2 + 2 \frac{\delta}{2\lambda_1} \xi_1 + \frac{\delta^2}{4\lambda_1^2} \right) - \frac{\delta^2}{4\lambda_1} + \epsilon \xi_2 + f

= \lambda_1 \left( \xi_1 + \frac{\delta}{2\lambda_1} \right)^2 + \epsilon \xi_2 + \left( f - \frac{\delta^2}{4\lambda_1} \right)

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

The conic section formed is a parabola:

\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.

Two-Dimensional Example

Conic sections in $\mathbb{R}^2$ satisfy the general quadratic equation:

ax_1^2 + bx_1x_2 + cx_2^2 + dx_1 + ex_2 + f = 0

Which can be written in matrix form as:

\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}^T \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} d \\ e \end{pmatrix}^T \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + f = 0

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

Which can be written in explicit form:

Q = \left\{ x \in \mathbb{R}^n : x^T A x + b^T x + c = 0 \right\}

= \left\{ x \in \mathbb{R}^n : \sum_{j=1}^n \sum_{k=1}^n x_j a_{jk} x_k + \sum_{j=1}^n b_j x_j + c = 0 \right\}

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 new coordinate basis $\xi = S^T x$ and $\mu = S^T b$ , the quadratic surface has diagonal form:

Q = \left\{ \xi \in \mathbb{R}^n : \xi^T S^T A S \xi + b^T S \xi + c = 0 \right\}

= \left\{ \xi \in \mathbb{R}^n : \xi^T \Lambda \xi + \mu^T \xi + c = 0 \right\}

= \left\{ \xi \in \mathbb{R}^n : \sum_{j=1}^n \lambda_j \xi_j^2 + \sum_{j=1}^n \mu_j \xi_j + c = 0 \right\}

= \left\{ \xi \in \mathbb{R}^n : \sum_{j=1}^n (\lambda_j \xi_j^2 + \mu_j \xi_j) = -c \right\}

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.

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