## 每週問題 January 31, 2011

Pow-Jan-31-11

### 4 Responses to 每週問題 January 31, 2011

1. levinc 說道：

直接分塊矩陣硬幹不知道行不行喔，還請老師開示
rk
(\begin{bmatrix}&space;O&space;&I&space;\\&space;I&space;&&space;-A&space;\end{bmatrix}
\begin{bmatrix}&space;A&space;&O&space;\\&space;I&space;&&space;B&space;\end{bmatrix}
\begin{bmatrix}&space;I&space;&-B&space;\\&space;O&space;&&space;I&space;\end{bmatrix})=
rk\begin{bmatrix}&space;I&space;&O&space;\\&space;O&space;&&space;-AB&space;\end{bmatrix}= n + rk(AB)\geq rkA +rkB. (由秩-零度定理)，及
rk
(\begin{bmatrix}&space;O&space;&I&space;\\&space;I&space;&&space;-AB^{-1}&space;\end{bmatrix}
\begin{bmatrix}&space;A&space;&O&space;\\&space;B&space;&&space;I&space;\end{bmatrix}
\begin{bmatrix}&space;I&space;&-B^{-1}I&space;\\&space;O&space;&&space;I&space;\end{bmatrix})
=
rk(\begin{bmatrix}&space;B&space;&O&space;\\&space;O&space;&&space;-AB^{-1}&space;\end{bmatrix})=rkB + rk(AB^{-1}).

所以咧，$rkB + rk(AB^{-1}) = n + rk(AB)\geq rkA +rkB$ ,
$rkB + rk(AB^{-1}) = \geq rkA +rkB iff rk(AB{-1})=rkA$, which provides that $rk(A)=rk(AB^{-1}B) = rk(AB)$. Done.

2. levinc 說道：

我的媽呀，亂七八糟 老師抱歉＞＜
$rk (\begin{bmatrix}&space;O&space;&I&space;\\&space;I&space;&&space;-A&space;\end{bmatrix} \begin{bmatrix}&space;A&space;&O&space;\\&space;I&space;&&space;B&space;\end{bmatrix} \begin{bmatrix}&space;I&space;&-B&space;\\&space;O&space;&&space;I&space;\end{bmatrix})= rk\begin{bmatrix}&space;I&space;&O&space;\\&space;O&space;&&space;-AB&space;\end{bmatrix}= n + rk(AB)\geq rkA +rkB.$ (由秩-零度定理)，及
$rk (\begin{bmatrix}&space;O&space;&I&space;\\&space;I&space;&&space;-AB^{-1}&space;\end{bmatrix} \begin{bmatrix}&space;A&space;&O&space;\\&space;B&space;&&space;I&space;\end{bmatrix} \begin{bmatrix}&space;I&space;&-B^{-1}I&space;\\&space;O&space;&&space;I&space;\end{bmatrix}) = rk(\begin{bmatrix}&space;B&space;&O&space;\\&space;O&space;&&space;-AB^{-1}&space;\end{bmatrix})=rkB + rk(AB^{-1}).$

所以咧，$rk(B)+rk(AB^{-1}) = n + rk(AB) \geq rkA + rkB$, 故
$rkB + rk(AB^{-1}) = rkA + rkB$ iff $rk(AB^{-1}) = rkA$, which provides that $rkA = rk(AB^{-1}B) = rk(AB)$. Done.

3. levinc 說道：

呃…蠢斃了….XDD
總之，中間那堆顯示不出來的部分就是在拆解分塊矩陣，
rk(左分塊)變成 rkB + rk(AB^{-1}), rk(右分塊)= n + rk(AB),
由已知，左右分塊秩相等，再用秩－零度定理，……然後接上面回嚮。

4. ccjou 說道：

如果 B 不是可逆矩陣…
可以試試先將分塊行列對調位置成容易判斷的形式，例如
$\mathrm{rk}\begin{bmatrix} A&0\\ B&I \end{bmatrix}=\mathrm{rk}\begin{bmatrix} O&A\\ I&B \end{bmatrix}=\mathrm{rk}\begin{bmatrix} I&B\\ 0&A \end{bmatrix}=\mathrm{rk}I+\mathrm{rk}A=n+\mathrm{rk}A$