「統計解析のための線形代数」復習筆記11

矩陣的定義

Theorem 1 (matrix) \(m\times n\) 個數 \(a_{ij} (i=1,2,\cdots,m; j=1,2,\cdots,n)\), 寫成縱 \(m\) 行, 橫 \(n\) 列的長方形或者正方形,左右用圓括號或者方括號包含在內。我們稱之爲 \(m\times n\) 矩陣(matrix),或者 \((m, n)\) 矩陣。 \(m\times n\) 或者 \((m,n)\) 被稱爲是這個矩陣的類型。我們常用大寫字母來標記一個矩陣,如下面的矩陣我們標記爲 \(A\)。 如果要特別明示矩陣的類型,可以寫作 \(\mathop{A}_{m\times n}, \mathop{A}_{(m, n)}, \; A(m\times n)\)。兩個矩陣如果行數相等,列數也相等,我們稱他們爲類型相同的矩陣。構成矩陣的一個個數 \(a_{11},a_{12},\cdots,a_{mn}\) 被叫做矩陣的成分(component, element, entry)。

\(i\)行,第\(j\)列交叉的地方的成分,\(a_{ij}\) 被叫做 \((i,j)\) 成分。矩陣有時候也會寫成 \(A=(a_{ij})\)

\(\left( \begin{array}{c} a_{11} & a_{12} & \cdots & a_{1j} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2j} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{i1} & a_{i2} & \cdots & a_{ij} & \cdots & a_{in}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mj} & \cdots & a_{mn}\\ \end{array} \right), \\ \left[ \begin{array}{c} a_{11} & a_{12} & \cdots & a_{1j} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2j} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{i1} & a_{i2} & \cdots & a_{ij} & \cdots & a_{in}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mj} & \cdots & a_{mn}\\ \end{array} \right]\)

  • 矩陣 \(\mathop{A}_{m\times n}\) 可以被看做是:
    以第\(1\)行爲成分的行向量 \((a_{11},a_{12},\cdots,a_{1n})=\underline{b}_1^t\)
    以第\(2\)行爲成分的行向量 \((a_{21},a_{22},\cdots,a_{2n})=\underline{b}_2^t\)
    \(\vdots\)
    以第 \(m\) 行爲成分的行向量 \((a_{m1},a_{m2},\cdots,a_{mn})=\underline{b}_m^t\)
    爲成分組成的列向量:
    \(\left( \begin{array}{c} \underline{b}_1^t\\ \underline{b}_2^t\\ \vdots\\ \underline{b}_m^t\\ \end{array} \right)\)

  • 類似的,矩陣 \(\mathop{A}_{m\times n}\) 可以被看做是:
    以第\(1\)列爲成分的列向量: \(\left( \begin{array}{c} a_{11}\\ a_{21}\\ \vdots\\ a_{m1}\\ \end{array} \right)=\underline{c}_1\)
    以第\(2\)列爲成分的列向量:\(\left( \begin{array}{c} a_{12}\\ a_{22}\\ \vdots\\ a_{m2}\\ \end{array} \right)=\underline{c}_2\)
    \(\vdots\)
    以第\(n\)列爲成分的列向量:\(\left( \begin{array}{c} a_{1n}\\ a_{2n}\\ \vdots\\ a_{mn}\\ \end{array} \right)=\underline{c}_n\)
    爲成分組成的行向量:\((\underline{c}_1,\underline{c}_2,\cdots,\underline{c}_n)\)

矩陣的運算,和零矩陣

矩陣的和與差

Theorem 2 (matrix plus or minus) 類型(type)相同的矩陣之間的加減法運算,被定義爲各個對應成分的加減法結果作成分的矩陣。

對於\(A=\left( \begin{array}{c} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{array} \right),\\ B=\left( \begin{array}{c} b_{11} & b_{12} & \cdots & b_{1n}\\ b_{21} & b_{22} & \cdots & b_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ b_{m1} & b_{m2} & \cdots & b_{mn}\\ \end{array} \right)\)
有:\(A\pm B=\left( \begin{array}{c} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{array} \right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\pm \left( \begin{array}{c} b_{11} & b_{12} & \cdots & b_{1n}\\ b_{21} & b_{22} & \cdots & b_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ b_{m1} & b_{m2} & \cdots & b_{mn}\\ \end{array} \right)\\ \;\;\;\;\;\;\;\;\;\;=\left( \begin{array}{c} a_{11}\pm b_{11} & a_{12}\pm b_{12} & \cdots & a_{1n}\pm b_{1n}\\ a_{21}\pm b_{21} & a_{22}\pm b_{22} & \cdots & a_{2n}\pm b_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1}\pm b_{m1} & a_{m2}\pm b_{m2} & \cdots & a_{mn}\pm b_{mn}\\ \end{array} \right)(復号同順)\)

  1. \(A=\left( \begin{array}{c} 9 & 3 & 1\\ -2 & 5 & 8\\ \end{array} \right), B=\left( \begin{array}{c} 4 & 2 & 1\\ 3 & -3 & 5\\ \end{array} \right)\) 那麼
    \(A+B = \left( \begin{array}{c} 9+4 & 3+2 & 1+1\\ -2+3 & 5+(-3) & 8+5\\ \end{array} \right)=\left( \begin{array}{c} 13 & 5 & 2\\ 1 & 2 & 13\\ \end{array} \right)\)

  2. 用1.中的矩陣運算:
    \(A-B=\left( \begin{array}{c} 9-4 & 3-2 & 1-1\\ -2-3 & 5-(-3) & 8-5\\ \end{array} \right)=\left( \begin{array}{c} 5 & 1 & 0\\ -5 & 8 & 3\\ \end{array} \right)\)

矩陣的相等

Theorem 3 (matrix equal) 類型相同的兩個矩陣 \(A,B\),如果他們對應的所有成分,一一相等,我們說這兩個矩陣是相等的。即:\(A=B\)

對於\(A=\left( \begin{array}{c} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{array} \right),\\ B=\left( \begin{array}{c} b_{11} & b_{12} & \cdots & b_{1n}\\ b_{21} & b_{22} & \cdots & b_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ b_{m1} & b_{m2} & \cdots & b_{mn}\\ \end{array} \right)\)
如果有:
\(a_{11}=b_{11},a_{12}=b_{12},\cdots,a_{mn}=b_{mn}\)
那麼 \(A=B\)

零矩陣

Theorem 4 (zero matrix) 所有的成分均爲數字 \(0\)\(m\times n\) 矩陣,
(共有 \(m\times n\) 個零。)
\(\left( \begin{array}{c} 0 & 0 & \cdots & 0\\ 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 0\\ \end{array} \right)\)
被稱爲零矩陣(zero matrix, null matrix)。寫作:\(\large 0, \mathop{\large 0}_{m\times n}, \mathop{\large 0}_{(m,n)}\)。要注意與標量的 \(0\) 區分。

矩陣的標量倍數運算

Theorem 5 (scalar times) 矩陣 \(A\) 的所有的成分,均乘以一個標量 \(k\),獲得新的矩陣的過程被稱爲矩陣的標量倍數運算。 寫作 \(kA\)

對於 \(A=\left( \begin{array}{c} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{array} \right)\)
有:\(kA = k\left( \begin{array}{c} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{array} \right)\\ =\left( \begin{array}{c} ka_{11} & ka_{12} & \cdots & ka_{1n}\\ ka_{21} & ka_{22} & \cdots & ka_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ ka_{m1} & ka_{m2} & \cdots & ka_{mn}\\ \end{array} \right)\)

特別的,當 \(k=-1\) 時, \((-1)A=-A\)\(k=0\) 時,\(0A=\Large 0\)。注意 \(\Large 0\) 是與 \(A\) 類型相同的零矩陣,而非標量 \(0\)

  1. \(A=\left( \begin{array}{c} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{array} \right)\)
    \(kA=\left( \begin{array}{c} ka_{11} & ka_{12} & ka_{13}\\ ka_{21} & ka_{22} & ka_{23}\\ ka_{31} & ka_{32} & ka_{33}\\ \end{array} \right)\)

  2. \(B=\left( \begin{array}{c} 1 & -2 & 3\\ -4 & 5 & -6\\ \end{array} \right)\)
    \(7B=\left( \begin{array}{c} 7\times1 & 7\times(-2) & 7\times3\\ 7\times(-4) & 7\times5 & 7\times(-6)\\ \end{array} \right)\\ \;\;\;\;=\left( \begin{array}{c} 7 & -14 & 21\\ -28 & 35 & -42\\ \end{array} \right)\)
    \(-B=\left( \begin{array}{c} -1 & 2 & -3\\ 4 & -5 & 6\\ \end{array} \right)\)
    \(0B=\left( \begin{array}{c} 0 & 0 & 0\\ 0 & 0 & 0\\ \end{array} \right)=\mathop{\large 0}_{2\times3}\)

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