「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) 將\(m\times n\) 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)。
第\(i\)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
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：\(A\pm B=\left( \begin{array}{c} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{array} \right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\pm \left( \begin{array}{c} b_{11} & b_{12} & \cdots & b_{1n}\\ b_{21} & b_{22} & \cdots & b_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ b_{m1} & b_{m2} & \cdots & b_{mn}\\ \end{array} \right)\\ \;\;\;\;\;\;\;\;\;\;=\left( \begin{array}{c} a_{11}\pm b_{11} & a_{12}\pm b_{12} & \cdots & a_{1n}\pm b_{1n}\\ a_{21}\pm b_{21} & a_{22}\pm b_{22} & \cdots & a_{2n}\pm b_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1}\pm b_{m1} & a_{m2}\pm b_{m2} & \cdots & a_{mn}\pm b_{mn}\\ \end{array} \right)(復号同順)\)
\(A=\left( \begin{array}{c} 9 & 3 & 1\\ -2 & 5 & 8\\ \end{array} \right)， B=\left( \begin{array}{c} 4 & 2 & 1\\ 3 & -3 & 5\\ \end{array} \right)\) 那麼
\(A+B = \left( \begin{array}{c} 9+4 & 3+2 & 1+1\\ -2+3 & 5+(-3) & 8+5\\ \end{array} \right)=\left( \begin{array}{c} 13 & 5 & 2\\ 1 & 2 & 13\\ \end{array} \right)\)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
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
(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
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 \(0\)。
對 \(A=\left( \begin{array}{c} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\\ \end{array} \right)\)，
\(kA=\left( \begin{array}{c} ka_{11} & ka_{12} & ka_{13}\\ ka_{21} & ka_{22} & ka_{23}\\ ka_{31} & ka_{32} & ka_{33}\\ \end{array} \right)\)對 \(B=\left( \begin{array}{c} 1 & -2 & 3\\ -4 & 5 & -6\\ \end{array} \right)\)，
\(7B=\left( \begin{array}{c} 7\times1 & 7\times(-2) & 7\times3\\ 7\times(-4) & 7\times5 & 7\times(-6)\\ \end{array} \right)\\ \;\;\;\;=\left( \begin{array}{c} 7 & -14 & 21\\ -28 & 35 & -42\\ \end{array} \right)\)
\(-B=\left( \begin{array}{c} -1 & 2 & -3\\ 4 & -5 & 6\\ \end{array} \right)\)；
\(0B=\left( \begin{array}{c} 0 & 0 & 0\\ 0 & 0 & 0\\ \end{array} \right)=\mathop{\large 0}_{2\times3}\)
