The end of the semester is approaching, and many top students are starting to get busy. However, for a non-top student like me, every day is basically the same. I hope to continue researching the things I am interested in deeply, because I believe that truly learning something useful is what matters most. Both my Mathematical Analysis and Higher Algebra teachers required course papers. I have written about "Eulerian Mathematics" and "Research on Hypercomplex Numbers," which I am quite interested in, and I will share that content with everyone later.
Although the semester is nearing its end, our classes are not yet finished. In fact, our new lessons continue until the eighteenth week. As exams approach, our "Higher Algebra" course is also coming to a close. Recently, we have been covering quadratic forms, specifically positive definite quadratic forms and positive definite matrices. Regarding the identification of positive definite matrices, textbooks provide two methods: one based on elementary transformations of the definition, and the other being the principal minor method (Sylvester’s Criterion). The former is beyond reproach, but I find the latter difficult to appreciate—while it is correct, it is "ugly" and computationally intensive. I haven’t yet figured out what is so good about the principal minor method. In my view, the method based on the discriminant of a quadratic equation discussed in this article is simpler and faster.
Positive Definite Quadratic Forms
A so-called positive definite quadratic form is a homogeneous quadratic function of n variables x_1, x_2, \dots, x_n that is always positive as long as the x_i are not all zero. For example: 2 x_1^2+x_2^2-2 x_1 x_2=x_1^2+(x_2-x_1)^2 This is a relatively simple positive definite quadratic form. A multivariate example would be: 5 x_1^2+x_2^2+5 x_3^2+4 x_1 x_2-8 x_1 x_3-4 x_2 x_3 and so on. However, it is not easy to see the positive definiteness of the above expression at a glance. To judge it, we could rewrite it in matrix form and then use elementary transformations or the principal minor method. But I judge it by considering the solutions to the equation: f(t)=5 x_1^2+x_2^2+5 t^2+4 x_1 x_2-(8 x_1 + 4 x_2 )t=0 First, it is easy to verify: f(0)=5 x_1^2+x_2^2+4 x_1 x_2=(2 x_1+x_2)^2 +x_1^2 \geq 0 Because its discriminant is: \begin{aligned} \Delta &=(8 x_1 + 4 x_2 )^2-4\times 5\times(5 x_1^2+x_2^2+4 x_1 x_2)\\ &=-36 x_1^2-4 x_2^2-16 x_1 x_2\\ &=-4[5 x_1^2+(2 x_2 -x_1)^2] \end{aligned} When x_1, x_2 are not both zero, the discriminant is negative; therefore, the original equation has no real roots, and f(t) is always greater than 0. When x_1=x_2=0, there is only one solution t=0. For t > 0, we also have f(t) > 0. Therefore, the quadratic form above is positive definite.
Positive Definite Matrices
By systematically organizing the above method, we obtain the following criterion for positive definite matrices.
A positive definite matrix is an n \times n real matrix \boldsymbol{A}_n such that for any non-zero n-dimensional real column vector \boldsymbol{x}, we have: \boldsymbol{x}^{T} \boldsymbol{A}_n \boldsymbol{x} > 0
Write \boldsymbol{A}_n in block form: \boldsymbol{A}_n=\left[ \begin{matrix} \boldsymbol{A}_{n-1}& \boldsymbol{b}\\ {\boldsymbol{c}^T}&d \end{matrix}\right] where \boldsymbol{A}_{n-1} is an (n-1) \times (n-1) matrix. Also, write the vector \boldsymbol{x} as a partitioned vector: \left( \begin{matrix} {\boldsymbol{y}}\\t \end{matrix} \right) where \boldsymbol{y} is an (n-1)-dimensional vector consisting of the first n-1 elements of \boldsymbol{x}, and t is the n-th element of \boldsymbol{x}. Then: \begin{aligned} f(t)=\boldsymbol{x}^T \boldsymbol{A}_n \boldsymbol{x}&=\left(\boldsymbol{y}^{T},t\right)\left[ \begin{array}{cc} \boldsymbol{A}_{n-1}& \boldsymbol{b}\\ {\boldsymbol{c}^T}&d \end{array} \right] \left( \begin{array}{c} {\boldsymbol{y}}\\ t \end{array} \right)\\ &=d t^2+\left(\boldsymbol{c}^{T} \boldsymbol{y}+\boldsymbol{y}^{T}\boldsymbol{b}\right)t+\boldsymbol{y}^{T}\boldsymbol{A}_{n-1}\boldsymbol{y}\\ &=d t^2+\left[\left(\boldsymbol{b}+\boldsymbol{c}\right)^{T}\boldsymbol{y}\right]t+\boldsymbol{y}^{T}\boldsymbol{A}_{n-1}\boldsymbol{y} \end{aligned}
If the matrix \boldsymbol{A}_n is positive definite, then f(0)=\boldsymbol{y}^{T}\boldsymbol{A}_{n-1}\boldsymbol{y} \geq 0, meaning \boldsymbol{A}_{n-1} must be positive definite; this is a necessary condition. Another necessary condition is d > 0.
Simultaneously, the discriminant of f(t)=0 must be non-positive: \begin{aligned} \Delta &=\left[\left(\boldsymbol{b}+\boldsymbol{c}\right)^{T}\boldsymbol{y}\right]^2-4d\boldsymbol{y}^{T}\boldsymbol{A}_{n-1}\boldsymbol{y}\\ &=4\boldsymbol{y}^{T}\left(\boldsymbol{m}\boldsymbol{m}^T-d\boldsymbol{A}_{n-1}\right)\boldsymbol{y}\\ &\leq 0 \end{aligned} where \boldsymbol{m}=\frac{1}{2}(\boldsymbol{b}+\boldsymbol{c}). This means that (d\boldsymbol{A}_{n-1}-\boldsymbol{m}\boldsymbol{m}^T) must be positive definite. Note that since the matrix \boldsymbol{m}\boldsymbol{m}^T is positive semi-definite, if (d\boldsymbol{A}_{n-1}-\boldsymbol{m}\boldsymbol{m}^T) is positive definite, then \boldsymbol{A}_{n-1} must be positive definite (as the sum of a positive definite matrix and a positive semi-definite matrix is positive definite), so the latter condition is already contained within the former. Thus, we obtain the following criterion:
A matrix \boldsymbol{A}_n=\left[ \begin{array}{cc} \boldsymbol{A}_{n-1}& \boldsymbol{b}\\ {\boldsymbol{c}^T}&d \end{array} \right] is positive definite if and only if d > 0 and (d\boldsymbol{A}_{n-1}-\boldsymbol{m}\boldsymbol{m}^T) is positive definite, where \boldsymbol{m}=\frac{1}{2}(\boldsymbol{b}+\boldsymbol{c}).
In this way, the problem of the positive definiteness of an n-th order matrix is transformed into the positive definiteness of an (n-1)-th order matrix, which can similarly be transformed into an (n-2)-th order matrix... until it becomes a simple positivity check for a single number.
Conclusion
No matter how you look at it, the computational load of the criterion provided in this article should be lower than that of elementary transformations or the principal minor method, and the derivation is relatively simple. Therefore, I still don’t understand why this is hardly mentioned in textbooks or online. In fact, even when doing homework, I prefer to use the discriminant method rather than conventional methods to judge positive definiteness. For me, a method I thought of myself that can solve the problem is a good method. Even if I hadn’t encountered the matrix foundations of quadratic forms, I could still use the discriminant to judge whether a quadratic form is positive definite. Even if I were to investigate for several years, I doubt I would come up with a method as "ugly" as the principal minor method.
However, what might surprise readers is that the "ugly" principal minor method is essentially the same as the method in this article. We first check if the bottom-right element of the diagonal is positive, then reduce the matrix by one order, and check if the bottom-right element of the new matrix is positive, and so on... The bottom-right element to be checked at each step is actually equivalent to a principal minor. There is no fundamental difference between the two, but if one were actually programming this on a computer, the method in this article would clearly be slightly superior in terms of efficiency.
When reprinting, please include the original link: https://kexue.fm/archives/2195
For more details on reprinting, please refer to: Scientific Space FAQ