## 秩分解──目視行秩等於列秩

$\displaystyle A=\left[\!\!\begin{array}{ccrr} 1&2&1&1\\ 1&2&-1&-3\\ 1&2&0&-1 \end{array}\!\!\right]$

$\left[\!\!\begin{array}{ccrrcccc} 1&2&1&1&\vline&1&0&0\\ 1&2&-1&-3&\vline&0&1&0\\ 1&2&0&-1&\vline&0&0&1 \end{array}\!\!\right]\rightarrow\left[\!\!\begin{array}{cccrcccr} 1&2&0&-1&\vline&0&0&1\\ 0&0&1&2&\vline&1&0&-1\\ 0&0&0&0&\vline&1&1&-2 \end{array}\!\!\right]$

\begin{aligned} E\begin{bmatrix} A&I_3 \end{bmatrix}&=\begin{bmatrix} EA&EI_3 \end{bmatrix}=\begin{bmatrix} R&E \end{bmatrix}\end{aligned}

$R=\left[\!\!\begin{array}{cccr} 1&2&0&-1\\ 0&0&1&2\\ 0&0&0&0 \end{array}\!\!\right],~~~E=\left[\!\!\begin{array}{ccr} 0&0&1\\ 1&0&-1\\ 1&1&-2 \end{array}\!\!\right]$

$\displaystyle E^{-1}=\left[\!\!\begin{array}{crc} 1&1&0\\ 1&-1&1\\ 1&0&0 \end{array}\!\!\right]$

\displaystyle\begin{aligned} A&=E^{-1}R=\left[\!\!\begin{array}{c} 1\\ 1\\ 1 \end{array}\!\!\right]\begin{bmatrix} 1&2&0&-1 \end{bmatrix}+\left[\!\!\begin{array}{r} 1\\ -1\\ 0 \end{array}\!\!\right]\begin{bmatrix} 0&0&1&2 \end{bmatrix}+\left[\!\!\begin{array}{c} 0\\ 1\\ 0 \end{array}\!\!\right]\begin{bmatrix} 0&0&0&0 \end{bmatrix}\\ &=\left[\!\!\begin{array}{cr} 1&1\\ 1&-1\\ 1&0 \end{array}\!\!\right]\left[\!\!\begin{array}{cccr} 1&2&0&-1\\ 0&0&1&2 \end{array}\!\!\right]=XY.\end{aligned}

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### 2 Responses to 秩分解──目視行秩等於列秩

1. 計組小粉絲 says:

文中第一列：行空間的維數由線 “性” 獨立的行向量個數決定

• ccjou says:

謝謝，已訂正。