(It is recommended to have a basic foundation in linear algebra to read this article, or at least an understanding of the basic concepts in linear algebra.)
This semester is drawing to a close, and our "Analytic Geometry" course has reached the stage of simplifying quadratic curves. However, for students who have not studied linear algebra, using the formulas for rotation and translation of axes directly can lead to an overwhelming amount of calculation. Although the "invariant" method is somewhat simpler in terms of calculation, it often feels mysterious—one wonders where it comes from, and it requires memorizing a stack of formulas. In fact, with a foundation in linear algebra, these concepts become quite easy to understand. I strive for a unified method to solve the same type of problem—that is, a unified way to handle all quadratic forms—and naturally, I hope the calculation remains simple.
The General Model
A general quadratic form can be written as: x^T A x + 2 b^T x + c = 0
Where x and b are n-dimensional column vectors (with elements x_i and b_i), A is an n \times n square matrix (with elements a_{ij}), and c is a constant. Here, we only discuss the cases for n=2 and n=3. The process of simplifying a quadratic form can be reduced to the simplification of the matrix A.
General articles discussing this problem often expand the matrix A into an (n+1)-dimensional matrix \widetilde{A}, represented in block form as: \widetilde{A} = \begin{bmatrix} A & b \\ b^T & c \end{bmatrix}
And the vector x is expanded into an (n+1)-dimensional vector \widetilde{x}: \widetilde{x} = \begin{bmatrix} x \\ 1 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \\ 1 \end{bmatrix}
Thus, the quadratic form becomes: \widetilde{x}^T \widetilde{A} \widetilde{x} = 0
In this case, the problem becomes the simplification of the matrix \widetilde{A}. However, in my view, while this method is concise and elegant, the computational cost is relatively high. For example, a quadratic curve in a plane would generate a 3 \times 3 matrix, and finding eigenvalues would change from a quadratic equation to a cubic equation. Furthermore, calculating determinants becomes significantly more troublesome with each increase in dimension. In this context, the benefits of conciseness feel somewhat "not worth the cost." Therefore, I attempt to simplify the n-dimensional quadratic form directly using n \times n matrices, yielding the following results.
My Idealized Model
Consider an isometry (distance-preserving transformation) matrix R. Being distance-preserving means that under this transformation, the "distance" of a vector remains unchanged. That is, v^T v = (Rv)^T (Rv) = v^T (R^T R) v, so we have R^T R = I, meaning the transpose of the matrix equals its inverse. Under such a transformation, the shape of the curve remains unchanged. The most common isometry is a rotation centered at the origin.
If the simplified quadratic form is of the form: x^T M x = k
Where M is a diagonal matrix and k is a constant, this will be the simplest form of the quadratic form and the most convenient for our research.
Consider the composition of an isometry and a translation, namely: x \to R(x + x_0)
Where x_0 is also an n-dimensional constant column vector. Substituting this into the equation above, we get: \begin{aligned} (x + x_0)^T R^T A R (x + x_0) &= k \\ x^T (R^T A R) x + x_0^T [(R^T A R) + (R^T A R)^T] x + x_0^T (R^T A R) x_0 - k &= 0 \end{aligned}
Corresponding to the original general quadratic form, we have: \begin{aligned} R^T M R &= A \\ [(R^T M R) + (R^T M R)^T] x_0 &= 2b \\ x_0^T (R^T M R) x_0 - k &= c \end{aligned}
Where R^T A R = R^{-1} A R = M is a similarity transformation. Generally, its simplest form is a diagonal matrix composed of all the eigenvalues of matrix A. Therefore, it is relatively easy to find its form. Next, we have: \begin{aligned} (A + A^T) x_0 &= 2b \\ x_0^T A x_0 - k &= c \end{aligned}
The key is to find the specific form of x_0, which is equivalent to solving a system of n linear equations: x_0 = 2(A + A^T)^{-1} b
Regarding quadratic forms, there are some very exciting results (refer to The Mathematical Bridge for proofs):
1. Any real quadratic form has a corresponding real symmetric matrix A;
2. Any real symmetric matrix A always has n real eigenvalues.
With these two guarantees, we don’t have to worry about cases where "eigenvalues" are complex numbers. Furthermore, the above calculation can be simplified further because for a symmetric matrix A^T = A, so it becomes: x_0 = A^{-1} b
General Treatment of Special Cases
At this point, the problem seems entirely solved. One just needs to find the eigenvalues first to obtain matrix M, then find x_0 = A^{-1} b, and finally substitute it into x_0^T A x_0 - k = c to find the constant term. Especially for quadratic curves, the amount of calculation is truly minimal. But the problem is, what if matrix A is not invertible? This is possible, specifically when an eigenvalue is 0, which for a plane corresponds to the parabolic case. In such a case, finding the inverse matrix is impossible. Does the method fail here?
For a parabola, we can still view it as the limit of a hyperbola or an ellipse. For example, the parabola 2py + x^2 = 0 can be seen as 2py + x^2 + \varepsilon y^2 where \varepsilon \to 0. Completing the square: x^2 + \varepsilon \left(y + \frac{p}{\varepsilon}\right)^2 - \frac{p^2}{\varepsilon} = 0 It still takes the form where all terms are squared.
This provides us with an idea: we can add a tiny deviation to the originally non-invertible matrix A to make it invertible. After finding the result, we then take the limit as the deviation approaches 0. This is a quite interesting approach. Let us explore further along this line.
The key is: where should the deviation be added? And how? Generally speaking, changing any element of the original matrix would work, but for the sake of convenience and uniformity, we decide to change the diagonal elements. If there is only one zero root, we add \varepsilon_1 to the first element on the diagonal of matrix A; if there are two zero roots, we continue by adding \varepsilon_2 to the second element on the diagonal, and so on. The purpose of this is to handle high-dimensional cases uniformly. Let the new matrix be \bar{A}. If there are k zero roots, then the determinant \det \bar{A} \approx (a_{11} + a_{22} + \dots + a_{nn}) \varepsilon_1 \varepsilon_2 \dots \varepsilon_k (i.e., the sum of the diagonal elements multiplied by all \varepsilon_i; readers can easily derive this from the knowledge of the "trace" of a matrix).
Thus, everything returns to the original steps:
The original n eigenvalues \lambda_i become \lambda_i - \varepsilon_i, so no zero roots appear (place the zero roots at the beginning). Then, find the constant vector x_0 via x_0 = (\bar{A})^{-1} b. Of course, this is an infinitely large vector, but that is not important; just keep the first-order infinitesimal terms. Then we can find the constant term: k = x_0^T A x_0 - c. It should take the following form: \frac{C_1}{\varepsilon_1} + \frac{C_2}{\varepsilon_2} + \dots + \text{Const}
Comparing this with: x^2 + \varepsilon \left(y + \frac{p}{\varepsilon}\right)^2 - \frac{p^2}{\varepsilon} = 0 It is not hard to find that 2\sqrt{C_i}x_i is the term corresponding to the eigenvalue that was 0 (the parabolic term).
Examples:
1. Elliptic (Hyperbolic) Type. x^2 + 6xy + y^2 + 6x + 2y - 1 = 0
Solution: The above equation is: \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + 2[3, 1] \begin{bmatrix} x \\ y \end{bmatrix} - 1 = 0 That is: A = \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}, \quad b = \begin{bmatrix} 3 \\ 1 \end{bmatrix}, \quad c = -1
Find the two eigenvalues of A: (1-\lambda)^2 - 3^2 = 0, which are -2 and 4.
Next, calculate: \begin{aligned} x_0 &= \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \\ k &= x_0^T A x_0 - c = x_0^T b - c = [0, 1] \begin{bmatrix} 3 \\ 1 \end{bmatrix} + 1 = 2 \end{aligned}
The final result is: \begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 2 Which is: -x^2 + 2y^2 = 1.
2. Parabolic Type x^2 - 4xy + 4y^2 + 2x - 2y - 1 = 0
Solution: The above equation is: \begin{bmatrix} 1 & -2 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + 2[1, -1] \begin{bmatrix} x \\ y \end{bmatrix} - 1 = 0 That is: A = \begin{bmatrix} 1 & -2 \\ -2 & 4 \end{bmatrix}, \quad b = \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \quad c = -1
The two eigenvalues are 0 and 5.
Change matrix A to \bar{A}: \bar{A} = \begin{bmatrix} 1+\varepsilon & -2 \\ -2 & 4 \end{bmatrix}
For a 2 \times 2 matrix, the inverse operation is simple: \begin{bmatrix} a & c \\ b & d \end{bmatrix}^{-1} = \frac{1}{\det A} \begin{bmatrix} d & -c \\ -b & a \end{bmatrix}
The approximate expansion of a 2 \times 2 determinant is also simple: \det \begin{bmatrix} a+\varepsilon & c \\ b & d \end{bmatrix} = ad - bc + d\varepsilon (Note: In this specific case, the trace-based approximation mentioned earlier gives 5\varepsilon).
Therefore: \bar{A}^{-1} = \frac{1}{5\varepsilon} \begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix} (Accurate to the first-order infinitesimal).
x_0 = \bar{A}^{-1} b = \frac{1}{5\varepsilon} \begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \frac{1}{5\varepsilon} \begin{bmatrix} 2 \\ 1 \end{bmatrix}
Calculate: k = x_0^T A x_0 - c \approx x_0^T b - c = \frac{1}{5\varepsilon} [2, 1] \begin{bmatrix} 1 \\ -1 \end{bmatrix} + 1 = \frac{1}{5\varepsilon} + 1
Correspondingly, this generates the 2\sqrt{\frac{1}{5}}y term. Thus, the quadratic curve is simplified to: 5x^2 \pm 2\sqrt{\frac{1}{5}}y = 0 (The sign here is not important; it just indicates whether it opens upward or downward).
For three-dimensional surfaces or even higher-dimensional cases, this method proceeds with the same steps. Although the computational load increases, the essence remains unchanged. I believe the "invariant" method of simplification is actually derived through a series of steps like those above. In analytic geometry, it is simply stripped of its linear algebra "clothing," and (perhaps unintentionally) given a somewhat confusing color.
At the same time, I sincerely hope that in analytic geometry examinations, problems should be posed without restricting the method. Do not present rigid problems like "simplify using the invariant method..." This does nothing to improve our abilities. After all, mastering the methods in the textbook is not the necessity; mastering the mathematical skills to handle a particular problem is the fundamental goal.
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