分块矩阵的初等变换不改变矩阵的秩
- 某一行左乘一个可逆矩阵
- 某一行左乘一个矩阵加到另一行
- 某两行交换位置
- 某一列右乘一个可逆矩阵
- 某一列右乘一个矩阵加到另一行
- 某两列交换位置
其中 2,5 还不改变矩阵的行列式.
利用 2,5 可以证明下面的打洞原理
打洞原理
\begin{array}{ll}\begin{vmatrix}A& B \\ C&D \end{vmatrix} &= \begin{vmatrix} A& B\\ 0 & D-CA^{-1} B \end{vmatrix}\end{array}
矩阵的秩
矩阵的秩有以下这些结论
$$rank\begin{pmatrix}A&0\\ 0&B\\ \end{pmatrix} = rank(A) + rank(B)$$
$$rank(A) + rank(B)\le rank\begin{pmatrix}A&C\\ 0&B\\ \end{pmatrix} = rank(A) + rank(B)+rank(C)$$
证明: 显然
西尔维斯不等式
$$rank(A)+rank(B)- n \le rank(AB)$$
分块矩阵秩相关不等式的基础
$$rank\ A \le \begin{pmatrix} A\\ B \end{pmatrix} , rank \ A \le rank (A,B)$$
通过这个不等式, 可以直接得到下面这个不等式
\begin{array}{ll}rank(A+B) &\le rank \begin{pmatrix}A \\ A+B\end{pmatrix}\\ &= rank \begin{pmatrix}A\\B\end{pmatrix}\\ &\le rank \begin{pmatrix} A&0\\ B& B\\ \end{pmatrix} \\ &= rank \begin{pmatrix}A&0\\ 0& B\\ \end{pmatrix} \\ &= rank \ A + rank \ B\end{array}
\begin{array}{ll} rank(A,B) &\le rank \begin{pmatrix} A& B\\ 0&B \end{pmatrix} \\ &= rank \begin{pmatrix} A& 0\\ 0&B \end{pmatrix}\\ &= rank(A) + rank(B) \end{array}
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Bodrum Escort Ve Masaj Hizmetleri
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