0 前言
线性代数中涉及到的一个非常重要的内容是二次型，而二次型中涉及到的一个非常重要的内容是二次型化标准型。也许很多人对二次型化标准型非常熟悉，但是如果问到二次型化标准型有什么用，可能挺多人会愣一愣。这篇文章给大家带来的就是二次型的一个应用——最值求解。
1 二次型化标准型
下面先简单回顾下二次型、标准型、二次型化标准型
1.1 二次型
二次型是指含有 n 个变量 \(x_1,x_2,\cdots,x_n\) 的二次齐次多项式，即在一个多项式中，未知数的个数为任意多个，但每一项的次数都为 2。它可以表示为如下形式：
\(f(x_1,x_2,\cdots,x_n) = a_{11}{x_1}^2 + a_{22}{x_2}^2 + \cdots + a_{nn}{x_n}^2 + a_{12}x_1x_2 + a_{13}x_1x_3 + \cdots + a_{n-1,n}x_{n-1}x_n\\\)或者更一般地表示为矩阵形式：
\(f(x) = x^TAx\\\)其中，\(x = {(x_1,x_2, \cdots,x_n)}^T\) 是变量向量，\(A = {(a_{ij})}_{n \times n}\) 是二次型矩阵，且由于二次型中的系数满足 \(a_{ij} = a_{ji}\)，所以A是实对称矩阵。
1.2 标准型
标准型是二次型的一种特殊形式，它只含有平方项，即形如：
\(f(x_1,x_2,\cdots,x_n) = \lambda_1y_1^2 + \lambda_2y_2^2 + \cdots + \lambda_ny_n^2\\\)其中，\(\lambda_i (i =1,2,\cdots,n)\) 是常数，且 \(y_1,y_2,\cdots,y_n\) 是原变量 \(x_1,x_2, \cdots,x_n\) 的线性变换。在标准型中，每一项都是某个变量的平方，没有交叉项。
1.3 二次型化标准型
二次型化标准型是指通过适当的坐标变换（如正交变换或可逆线性变换），将一般的二次型转化为标准型的过程。这个过程在理论和实际应用中都具有重要意义。
具体步骤通常包括：
- 求二次型矩阵 \(A\) 的特征值和特征向量：由于 \(A\) 是实对称矩阵，它一定可以相似对角化，即存在正交矩阵 \(Q\)，使得 \(Q^TAQ = \Lambda\)，其中 \(\Lambda\) 是对角矩阵，对角线上的元素是 \(A\) 的特征值。
- 进行坐标变换：令 \(y = Q^Tx\)，则原二次型可以表示为：
\(\\f(x) = x^TAx = (Qy)^TA(Qy) = y^T(Q^TAQ)y = y^T\Lambda y = \lambda_1y_1^2 + \lambda_2y_2^2 + \cdots + \lambda_ny_n^2\)
这样就将原二次型化为了标准型。
2 极值求解
下面来看看二次型化标准型如何帮助我们计算最值。
假设现在有这样一个二次型
\(f(x) = x^TAx\\\)其中，\(x = {(x_1,x_2, \cdots,x_n)}^T\) 是变量向量，\(A = {(a_{ij})}_{n \times n}\) 是二次型矩阵。
2.1 化二次型
由于 \(A\) 是实对称矩阵，它一定可以相似对角化，即存在正交矩阵 \(Q\)，使得 \(Q^TAQ = \Lambda\)。我们令 \(y = Q^Tx\)，则有
\(\\f(x) = x^TAx = (Qy)^TA(Qy) = y^T(Q^TAQ)y = y^T\Lambda y = \lambda_1y_1^2 + \lambda_2y_2^2 + \cdots + \lambda_ny_n^2\)
2.2 观察总结
化完二次型后，接下来求最值就简单很多了。我们仔细看下面这个式子
\(f(x) = \lambda_1y_1^2 + \lambda_2y_2^2 + \cdots + \lambda_ny_n^2\)如果我们想要求这个式子的最值，\(\lambda_i(i=1,2,\cdots,n)\) 的取值是非常重要的。
2.2.1 特殊情况
我们先来从几种特殊情况入手。
第一种情况，如果对于所有的 \(i = 1,2,\cdots,n\)，都有 \(\lambda_i\) 大于 0，那么，无论 \(y_i (i=1,2,\cdots,n)\) 如何取值，都有 \(f(x) \ge 0\)，特别地，当对于所有的 \(i = 1,2,\cdots,n\)，\(y_i = 0\)，那么此时取得最小值 \(f(x) = 0\)。
第二种情况，如果对于所有的 \(i = 1,2,\cdots,n\)，都有 \(\lambda_i\)小于 0，那么，无论 \(y_i (i=1,2,\cdots,n)\) 如何取值，都有 \(f(x) \le 0\)，特别地，当对于所有的 \(i = 1,2,\cdots,n\)，\(y_i = 0\)，那么此时取得最大值 \(f(x) = 0\)。
第三种情况，如果对于所有的 \(i = 1,2,\cdots,n\)，都有 \(\lambda_i\)等于 0，那么，无论 \(y_i (i=1,2,\cdots,n)\) 如何取值，都有 \(f(x) = 0\)。
好的，我们已经清楚了 \(\lambda_i(i=1,2,\cdots,n)\) 全为正、或全为负、或全为零的情况。
2.2.2 普通情况
现在我们来思考一种相对普通情况，假设 \(\lambda_i(i=1,2,\cdots,n)\) 中，既有正，又有负。
首先我想说，这种情况，没有极大值也没有极小值。我们可以从两个关键点来思考：
- 当 \(\lambda_i\)大于 0，项 \(\lambda_iy_i^2\) 随着 \(y_i\) 的增大而增大，这意味着在这个方向上，函数值可以变得无限大。
- 当 \(\lambda_i\)小于 0，项 \(\lambda_iy_i^2\) 随着 \(y_i\) 的增大而减小，这意味着在这个方向上，函数值可以变得无限小。
这意味着不存在最大值和最小值，因为你可以总是找到更大的 y 值使得函数值超越任何给定的界限。
写到这里，我想我们都能至少有些模糊的感受，特征值 \(\lambda_i\) 的正负性非常重要，我们可以发现：
当有 \(\lambda_i\)大于 0时，\(f(x)\) 可以无限大，没有最大值。当 \(\lambda_i\)小于 0 时，\(f(x)\) 可以无限小，没有最小值。
2.3 结论
以上过程，我想表达的是一个思考的过程，从特殊情况到相对普通情况的一个探索，至此，我们对于特征值 \(\lambda_i\) 建立了一些直观感受。下面我们来整理结论（我不会再进行推导，我相信你们思考出来）。给一个标准型，我们可以快速判断其最大值和最小值情况。
对于最大值：
\begin{cases}
无最大值 & \exists \lambda_i > 0 \\
最大值为 0，无论所有 y_i 如何取值 & \forall \lambda_i = 0 \\
最大值为 0，要求\lambda_i < 0 对应的 y_i 取 0，其他 y_i 任取 & 其他\ (\forall \lambda_i\le 0, \exists \lambda_i < 0)
\end{cases}
对于最小值：
\begin{cases}
无最小值 & \exists \lambda_i < 0 \\
最小值为 0，无论所有 y_i 如何取值 & \forall \lambda_i = 0 \\
最小值为 0，要求\lambda_i > 0 对应的 y_i 取 0，其他 y_i 任取 & 其他\ (\forall \lambda_i\ge 0, \exists \lambda_i > 0)
\end{cases}
上面的核心是要注意什么时候应该让 \(y_i\) 取 0，这是我们要思考和留意的点。
好了，到这里，我们已经知道标准型什么时候有最值，以及取得最值时 \(y\) 的取值。
进而，由于 \(y = Q^Tx\)，且 \(Q\) 是正交矩阵（可逆），因此，对于任意的 \(y\) 都有唯一对应的 \(x\)，于是我们便可以求出取得最值时 \(x\) 的取值。
以上，即为二次型化标准型对于求最值的帮助，学而有益，希望对大家有启发。
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