網友JERRY留言：
已知 
答曰：
我們先舉一些例子看會發生何種情況。若 
線性方程
給定 
![\displaystyle A=\begin{bmatrix} 1&2\\ 0&3 \end{bmatrix},~~B=\left[\!\!\begin{array}{rr} 5&-2\\ 4&-1 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++A%3D%5Cbegin%7Bbmatrix%7D++1%262%5C%5C++0%263++%5Cend%7Bbmatrix%7D%2C%7E%7EB%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brr%7D++5%26-2%5C%5C++4%26-1++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
代入上述方程式等號兩邊，乘開可得，
上面兩矩陣相等，即有
或以矩陣形式表示為
![\displaystyle \left[\!\!\begin{array}{rrrr} 4&4&-2&0\\ -2&-2&0&-2\\ 0&0&2&4\\ 0&0&-2&-4 \end{array}\!\!\right]\begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brrrr%7D++4%264%26-2%260%5C%5C++-2%26-2%260%26-2%5C%5C++0%260%262%264%5C%5C++0%260%26-2%26-4++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cbegin%7Bbmatrix%7D++a%5C%5C++b%5C%5C++c%5C%5C++d++%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D++0%5C%5C++0%5C%5C++0%5C%5C++0++%5Cend%7Bbmatrix%7D&bg=ffffff&fg=000000&s=0&c=20201002)
下一步要解出這個齊次方程式。使用高斯消去法化簡係數矩陣，過程如下：
![\displaystyle \left[\!\!\begin{array}{rrrr} 4&4&-2&0\\ -2&-2&0&-2\\ 0&0&2&4\\ 0&0&-2&-4 \end{array}\!\!\right]\to\left[\!\!\begin{array}{rrrr} 1&1&-1/2&0\\ 0&0&-1&-2\\ 0&0&2&4\\ 0&0&-2&-4 \end{array}\!\!\right]\to\left[\!\!\begin{array}{rrrr} 1&1&0&1\\ 0&0&1&2\\ 0&0&0&0\\ 0&0&0&0 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brrrr%7D++4%264%26-2%260%5C%5C++-2%26-2%260%26-2%5C%5C++0%260%262%264%5C%5C++0%260%26-2%26-4++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cto%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brrrr%7D++1%261%26-1%2F2%260%5C%5C++0%260%26-1%26-2%5C%5C++0%260%262%264%5C%5C++0%260%26-2%26-4++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cto%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brrrr%7D++1%261%260%261%5C%5C++0%260%261%262%5C%5C++0%260%260%260%5C%5C++0%260%260%260++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
我們得到等價的齊次方程組：
解出 
但
上式指出 
表面上，以線性方程解 
對角化
數學家發現 
當 
沿用前例說明計算過程。將二階矩陣
![\displaystyle \Lambda=\begin{bmatrix} 1&0\\ 0&3 \end{bmatrix},~S=\begin{bmatrix} 1&1\\ 0&1 \end{bmatrix},~S^{-1}=\left[\!\!\begin{array}{cr} 1&-1\\ 0&1 \end{array}\!\!\right],~T=\begin{bmatrix} 1&1\\ 2&1 \end{bmatrix},~T^{-1}=\left[\!\!\begin{array}{rr} -1&1\\ 2&-1 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5CLambda%3D%5Cbegin%7Bbmatrix%7D++1%260%5C%5C++0%263++%5Cend%7Bbmatrix%7D%2C%7ES%3D%5Cbegin%7Bbmatrix%7D++1%261%5C%5C++0%261++%5Cend%7Bbmatrix%7D%2C%7ES%5E%7B-1%7D%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcr%7D++1%26-1%5C%5C++0%261++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%2C%7ET%3D%5Cbegin%7Bbmatrix%7D++1%261%5C%5C++2%261++%5Cend%7Bbmatrix%7D%2C%7ET%5E%7B-1%7D%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brr%7D++-1%261%5C%5C++2%26-1++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
請注意，
![\displaystyle P=ST^{-1}=\begin{bmatrix} \gamma&\delta\\ 0&\delta \end{bmatrix}\left[\!\!\begin{array}{rr} -1&1\\ 2&-1 \end{array}\!\!\right]=\begin{bmatrix} -\gamma+2\delta&\gamma-\delta\\ 2\delta&-\delta \end{bmatrix}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++P%3DST%5E%7B-1%7D%3D%5Cbegin%7Bbmatrix%7D++%5Cgamma%26%5Cdelta%5C%5C++0%26%5Cdelta++%5Cend%7Bbmatrix%7D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brr%7D++-1%261%5C%5C++2%26-1++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%3D%5Cbegin%7Bbmatrix%7D++-%5Cgamma%2B2%5Cdelta%26%5Cgamma-%5Cdelta%5C%5C++2%5Cdelta%26-%5Cdelta++%5Cend%7Bbmatrix%7D&bg=ffffff&fg=000000&s=0&c=20201002)
將此 
最後還有一個問題要釐清。當
線代啟示錄 