矩阵¶
常用类型¶
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\(m \times n\) 矩阵
由 \(m \times n\) 个元素 \(a_{ij}\) 排列而成.
\[ \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} \]记作 \((a_{ij})\) 或 \((a_{ij})_{m \times n}\), 也可使用大写字母表示, 如 \(A_{m \times n}\).
- 特别地, \(m = n\), 即行列相等时, 矩阵称为 \(n\) 阶方阵
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零矩阵
矩阵元素 \(a_{ij}\) 皆为 \(0\) 的矩阵, 记作 \(O\).
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对角矩阵
除对角线外的元素均为 \(0\) 的矩阵称为对角矩阵, 记作 \(diag(a_1, a_2, \cdots, a_n)\), 即
\[ diag(a_1, a_2, \cdots, a_n) = \begin{pmatrix} a_1 & 0 & \cdots & 0 \\ 0 & a_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_n \end{pmatrix} \]-
特别地, 当 \(a_1 = a_2 = \cdots = a_n = 1\) 时, 称为 单位矩阵, 记作 \(E_n\):
\[ E_n = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix} \]
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行(列)矩阵
只有一行或一列的矩阵被称为行(列)矩阵/向量
\[ A = (a_1, a_2, \cdots, a_n) \]\[ B = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} \]
运算¶
线性运算¶
加法¶
Warning
只有两个矩阵为同型矩阵时, 才能进行加法运算.
直接把各个元素对应相加即可.
设有同型矩阵 \(A = (a_{ij})_{m \times n}\), \(B = (b_{ij})_{m \times n}\), 则
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矩阵加法满足以下运算律
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交换律: \(A + B = B + A\)
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结合律: \((A + B) + C = A + (B + C)\)
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\(A + O = A\)
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数乘¶
和加法类似, 数乘就是把矩阵的每个元素都乘上一个数.
设有矩阵 \(A = (a_{ij})_{m \times n}\), 数 \(\lambda\), 则有
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矩阵加法满足以下运算律
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交换率
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结合律: \((\lambda \mu) A = \lambda (\mu A)\)
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分配率:
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\((\lambda + \mu) A = \lambda A + \mu A\)
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\(\lambda (A + B) = \lambda A + \lambda B\)
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矩阵乘法¶
定义¶
矩阵乘法本质上是"行与列的对应关系".
有矩阵 \(A = (a_{ik})_{m \times t}\), \(B = (b_{kj})_{t \times n}\), 则有 \(C = (c_{ij})_{m \times n}\), 称为矩阵 \(A\) 左乘 矩阵 \(B\)(或 \(B\) 右乘 \(A\))之积, 记作
其中
Warning
这里注意 \(A\) 的行数与 \(B\) 的列数必须是一致的
只看公式可能不太好理解, 下面看几个例子
Example
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设 \(A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix}\), \(B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{pmatrix}\), 则有
\[ AB = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{pmatrix} = \begin{pmatrix} a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} & a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32} \\ a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} & a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32} \end{pmatrix} \] -
设 \(A = (a, b, c)\), \(B = \begin{pmatrix} d \\ e \\ f \end{pmatrix}\), 则有
\[ AB = (a, b, c) \begin{pmatrix} d \\ e \\ f \end{pmatrix} = ad + be + cf \]\[ BA = \begin{pmatrix} d \\ e \\ f \end{pmatrix} (a, b, c) = \begin{pmatrix} da & db & dc \\ ea & eb & ec \\ fa & fb & fc \end{pmatrix} \] -
求解 \((x_1, x_2, x_3) \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}\)
\[ \begin{align} \text{原式} & = (x_1, x_2, x_3) \begin{pmatrix} a_{11}x_1 + a_{12}x_2 + a_{13}x_3 \\ a_{21}x_1 + a_{22}x_2 + a_{23}x_3 \\ a_{31}x_1 + a_{32}x_2 + a_{33}x_3 \end{pmatrix} \\ & = x_1(a_{11}x_1 + a_{12}x_2 + a_{13}x_3) + x_2(a_{21}x_1 + a_{22}x_2 + a_{23}x_3) + x_3(a_{31}x_1 + a_{32}x_2 + a_{33}x_3) \\ & = a_{11}x_1^2 + a_{12}x_1x_2 + a_{13}x_1x_3 + a_{21}x_2x_1 + a_{22}x_2^2 + a_{23}x_2x_3 + a_{31}x_3x_1 + a_{32}x_3x_2 + a_{33}x_3^2 \\ & = a_{11}x_1^2 + a_{22}x_2^2 + a_{33}x_3^2 + (a_{12} + a_{21})x_1x_2 + (a_{13} + a_{31})x_1x_3 + (a_{23} + a_{32})x_2x_3 \\ \end{align} \]
简单来说就是, 左矩阵第 \(i\) 行的元素与右矩阵第 \(j\) 列对应元素相乘后求和，得到结果矩阵的第 \((i, j)\) 个元素, 左矩阵有几行, 结果就有几行; 右矩阵有几列, 结果就有几列.
运算律¶
从上面的第二个例子不难看出, 矩阵乘法不满足交换律.
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\((AB)C = A(BC)\)
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\((\lambda A)B = \lambda (AB)\)
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\(A(B + C) = AB + AC\)
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\((A + B)C = AC + BC\)
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对于单位矩阵 \(E_n\), 有
\[ A_{m \times n}E_{n} = A_{m \times n} \]\[ E_{n}A_{m \times n} = A_{m \times n} \]即
\[ AE = EA = A \]Example
设 \(A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix}\), \(E_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\), \(E_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\), 则有
\[ AE_3 = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} = A \]\[ E_2A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} = A \]
矩阵的幂¶
设 \(A\) 为 \(n\) 阶方阵, 则有:
矩阵的幂运算满足以下运算率 \((k \in \mathbb{Z}, n \in \mathbb{N^{+}})\):
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\(A^{n_1 + n_2} = A^{n_1} A^{n_2}\)
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\((A^{n_1})^{n_2} = A^{n_1 n_2}\)
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\((kA)^{n} = k^{n} A^{n}\)
Warning
由于矩阵乘法不满足交换律, 故有以下结论:
将左右两边分别展开对比一下就知道了:
不难发现中间两项 \(A\) \(B\) 相乘的顺序不同. 但且仅当矩阵 \(A\) \(B\) 可交换时, 方可像实数一样进行幂运算.
Example
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设 \(A = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\), \(B = \begin{pmatrix} 3 & 1 & -2 \end{pmatrix}\), 求 \((AB)^{n}\).
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解:
\[ AB = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \begin{pmatrix} 3 & 1 & -2 \end{pmatrix} = \begin{pmatrix} 3 & 1 & -2 \\ -3 & -1 & 2 \\ 6 & 2 & -4 \end{pmatrix} \]\[ BA = \begin{pmatrix} 3 & 1 & -2 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = 3 - 1 - 4 = -2 \]故有
\[ \begin{align} \text{原式} & = A \cdot \underbrace{BABABA}_{\text{n-1组}} \cdots B \\ & = (-2)^{n - 1} AB \\ & = (-2)^{n - 1} \begin{pmatrix} 3 & 1 & -2 \\ -3 & -1 & 2 \\ 6 & 2 & -4 \end{pmatrix} \end{align} \]
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设 \(A = \begin{pmatrix} 2 & 4 & -6 \\ 1 & 2 & -3 \\ 4 & 8 & -12 \end{pmatrix}\), 求 \(A^{100}\).
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解:
观察易发现, \(A\) 的行(列)成比例, \(R(n) = 1\)1
令 \(A = \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix} \begin{pmatrix} 1 & 2 & -3 \end{pmatrix} = \alpha \beta^{T}\)
则有
\[ \beta^{T} \alpha = \begin{pmatrix} 1 & 2 & -3 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix} = 2 + 2 - 12 = -8 \]故
\[ \begin{align} \text{原式} & = \alpha \cdot \underbrace{\beta^{T}\alpha\beta^{T}\alpha\beta^{T}\alpha}_{\text{99 组}} \cdots \beta^{T} \\ & = (-8)^{99} \alpha\beta^{T} \\ & = - 8^{99} \begin{pmatrix} 2 & 4 & -6 \\ 1 & 2 & -3 \\ 4 & 8 & -12 \end{pmatrix} \end{align} \]
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矩阵转置¶
转置后 \(A^{T}\) 的列数与 \(B^{T}\) 的行数不再相等, 无法再进行乘法运算, 但 \(B^{T}\) 的列数与 \(A^{T}\) 行数是相等的.
推广
方阵的行列式¶
定义¶
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只有方阵才有行列式
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行列式的本质是一个数
Example
设有方阵 \(A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & -1 & 4 \\ 0 & 0 & 5 \end{pmatrix}\), 则其行列式为 \(\begin{vmatrix} 1 & 2 & 3 \\ 0 & -1 & 4 \\ 0 & 0 & 5 \end{vmatrix} = -5\)
性质¶
设 \(A\), \(B\) 为 \(n\) 阶方阵, \(k\) 为常数, \(m\) 为正整数.
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\(|A^{T}| = |A|\)
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\(|kA| = k^{n} |A|\)
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\(|AB| = |A| \cdot |B|\)
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\(|A^{m}| = |A|^{m}\)
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\(|E| = 1\)
Example
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设 \(A\) 为 \(n\) 阶方阵, 且 \(|A| = 3\), 求 \(||A|A^{T}|\), \(||A|A^{2}|\)
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解:
\[ ||A|A^{T}| = |3A^{T}| = 3^{n}|A^{T}| = 3^{n}|A| = 3^{n} \cdot 3 = 3^{n + 1} \]\[ ||A|A^{2}| = |3A^{2}| = 3^{n}|A^{2}| = 3^{n}|A|^{2} = 3^{n}3^{2} = 3^{n + 2} \]
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已知 \(A = \begin{pmatrix} 2 & 1 \\ -1 & 2 \end{pmatrix}\), \(E\) 为 \(2\) 阶单位矩阵, 矩阵 \(B\) 满足 \(BA = B + 2E\), 求 \(|B|\)
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解:
依题可得 \(BA - BE = 2E\), 即 \(B(A -E) = 2E\), 两边同时取行列式, 得:
\[ |B(A - E)| = |2E| \]即
\[ |B| \cdot |A - E| = 2^{2}|E| = 4 \]又 \(A -E = \begin{pmatrix} 2 & 1 \\ -1 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}\)
故 \(|A - E| = 2\), 代入得 \(|B| \cdot 2 = 4\), 故 \(|B| = 2\).
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设 \(n\) 阶矩阵 \(A\) 满足 \(A^{T}A = E\), 其中 \(E\) 为 \(n\) 阶单位矩阵, 若 \(|A| < 0\), 求 \(|A + E|\)
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解:
\[ \begin{align} |A + E| & = |A + A^{T}A| \\ & = |EA + A^{T}A| \\ & = |(E + A^{T})A| \\ & = |E + A^{T}| \cdot |A| \\ & = - |E^{T} + A^{T}| \\ & = - |(E + A)^{T}| \\ & = - |E + A| \end{align} \]即 \(2|A + E| = 0\), 故 \(|A + E| = 0\)
错误解法
\(|A^{T}A| = |E| \Rightarrow |A^{T}| \cdot |A| = 1 \Rightarrow |A|^{2} = 1 \Rightarrow |A| = \pm 1\)
又因为 \(|A| < 0\), 故 \(|A| = -1\)
所以 \(|A + E| = |A| + |E| = -1 + 1 = 0\)
⚠️ ️️\(|A + B| \not ={|A| + |B|}\)
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