本文的閱讀等級：初級
矩陣的行空間的維數稱為行秩 (column rank)，列空間的維數稱為列秩 (row rank)。子空間的維數由最大的線性獨立的向量數決定，“行秩=列秩”一文曾基於此性質通過操作矩陣乘法運算證明了矩陣的行秩等於列秩。證明歸證明，讀者心中可能依然困惑：「矩陣的線性獨立行向量數怎麼會恰好等於線性獨立的列向量數呢？」本文再提供一個論證，想法很簡單：利用高斯消去法挑選出矩陣的線性獨立行與列，並以一個特殊分解式呈現獨立行與獨立列。這個證明屬計算導向，雖未直接表達行秩等於列秩的幾何特性，但由所得的矩陣分解式我們可以「目視」原矩陣的行空間和列空間，兩者確實擁有相等的基底向量數。
下面我用一個例子來說明計算過程。考慮 
![\displaystyle A=\left[\!\!\begin{array}{ccrr} 1&2&1&1\\ 1&2&-1&-3\\ 1&2&0&-1 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+A%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bccrr%7D++1%262%261%261%5C%5C++1%262%26-1%26-3%5C%5C++1%262%260%26-1++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
利用高斯消去法化簡增廣矩陣 
![\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]](https://s0.wp.com/latex.php?latex=%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bccrrcccc%7D++1%262%261%261%26%5Cvline%261%260%260%5C%5C++1%262%26-1%26-3%26%5Cvline%260%261%260%5C%5C++1%262%260%26-1%26%5Cvline%260%260%261++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Crightarrow%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcccrcccr%7D++1%262%260%26-1%26%5Cvline%260%260%261%5C%5C++0%260%261%262%26%5Cvline%261%260%26-1%5C%5C++0%260%260%260%26%5Cvline%261%261%26-2++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
消去過程可以解讀為一連串的基本矩陣乘法，總效果為
其中
![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]](https://s0.wp.com/latex.php?latex=R%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcccr%7D++1%262%260%26-1%5C%5C++0%260%261%262%5C%5C++0%260%260%260++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%2C%7E%7E%7EE%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bccr%7D++0%260%261%5C%5C++1%260%26-1%5C%5C++1%261%26-2++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
方陣 
![\displaystyle E^{-1}=\left[\!\!\begin{array}{crc} 1&1&0\\ 1&-1&1\\ 1&0&0 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+E%5E%7B-1%7D%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcrc%7D++1%261%260%5C%5C++1%26-1%261%5C%5C++1%260%260++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
觀察發現 
![\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}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D+A%26%3DE%5E%7B-1%7DR%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bc%7D++1%5C%5C++1%5C%5C++1++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cbegin%7Bbmatrix%7D+1%262%260%26-1+%5Cend%7Bbmatrix%7D%2B%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Br%7D++1%5C%5C++-1%5C%5C++0++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cbegin%7Bbmatrix%7D+0%260%261%262+%5Cend%7Bbmatrix%7D%2B%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bc%7D++0%5C%5C++1%5C%5C++0++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cbegin%7Bbmatrix%7D+0%260%260%260+%5Cend%7Bbmatrix%7D%5C%5C+%26%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcr%7D++1%261%5C%5C++1%26-1%5C%5C++1%260++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcccr%7D++1%262%260%26-1%5C%5C++0%260%261%262++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%3DXY.%5Cend%7Baligned%7D&bg=ffffff&fg=000000&s=0&c=20201002)
因為 
秩分解是等價標準型 (見“每週問題 April 20, 2009”) 的約化。秩分解具有哪些實際用途呢？雖然秩分解既不像 LU 分解或 QR 分解可以用來解線性方程，也不如奇異值分解 (SVD) 具備對應良好的正交基底，但秩分解常用於推導化簡，如表達 Moore-Penrose 偽逆矩陣 (見“Moore-Penrose 偽逆矩陣”)，並有展示行空間基底和列空間基底的作用。本文利用秩分解來證明線性代數的一個重要基石：矩陣的行秩等於列秩，也就是說，
線代啟示錄 
文中第一列：行空間的維數由線 “性” 獨立的行向量個數決定
謝謝，已訂正。