## 每週問題 October 27, 2014

$A$$B$ 是同尺寸矩陣，證明 $\text{rank}(A+B)\le\text{rank}A+\text{rank}B$

Let $A$ and $B$ be $m\times n$ matrices. Show that $\text{rank}(A+B)\le\text{rank}A+\text{rank}B$.

$\text{rank}(A+B)=\dim C(A+B)\le \dim (C(A)+C(B))$

$\{\mathbf{u}_1,\ldots,\mathbf{u}_r\}$$C(A)$ 的基底，$\{\mathbf{v}_1,\ldots,\mathbf{v}_s\}$$C(B)$ 的基底，其中 $r=\text{rank}A$$s=\text{rank}B$。據此，$C(A)+C(B)=\text{span}\{\mathbf{u}_1,\ldots,\mathbf{u}_r,\mathbf{v}_1,\ldots,\mathbf{v}_s\}$。明顯地，

$\dim (C(A)+C(B))\le r+s=\text{rank}A+\text{rank}B$

$\begin{bmatrix} I_m&I_m\\ 0&I_m \end{bmatrix}\begin{bmatrix} A&0\\ 0&B \end{bmatrix}\begin{bmatrix} I_n&0\\ I_n&I_n \end{bmatrix}=\begin{bmatrix} A+B&B\\ B&B \end{bmatrix}$

\begin{aligned} \hbox{rank}A+\hbox{rank}B&=\hbox{rank}\begin{bmatrix} A&0\\ 0&B \end{bmatrix}=\hbox{rank}\begin{bmatrix} A+B&B\\ B&B \end{bmatrix}\\ &\ge\hbox{rank}\begin{bmatrix} A+B\\ B \end{bmatrix} \ge\hbox{rank}(A+B). \end{aligned}

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