## 每週問題 December 14, 2015

Let $A$ and $B$ be $n\times n$ matrices. If $AB=BA$, show that

$\hbox{rank}(A+B)\le\hbox{rank}A+\hbox{rank}B-\hbox{rank}(AB)$.

\begin{aligned} C(A+B)&=\{A\mathbf{x}+B\mathbf{x}\vert\mathbf{x}\in\mathbb{C}^n\}\\ &\subseteq \{A\mathbf{x}\vert\mathbf{x}\in\mathbb{C}^n\}+\{B\mathbf{x}\vert\mathbf{x}\in\mathbb{C}^n\}\\ &=C(A)+C(B).\end{aligned}

\begin{aligned} \dim C(A+B)&\le \dim (C(A)+C(B))\\ &=\dim C(A)+\dim C(B)-\dim (C(A)\cap C(B)), \end{aligned}

$\hbox{rank}(A+B)\le \hbox{rank}A+\hbox{rank}B-\dim (C(A)\cap C(B))$

$\hbox{rank}(AB)\le \dim (C(A)\cap C(B))$

$\begin{bmatrix} I&I\\ 0&I \end{bmatrix}\begin{bmatrix} A&0\\ 0&B \end{bmatrix}\begin{bmatrix} -B&I\\ A&I \end{bmatrix}=\begin{bmatrix} AB-BA&A+B\\ BA&B \end{bmatrix}=\begin{bmatrix} 0&A+B\\ AB&B \end{bmatrix}$

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

### 2 Responses to 每週問題 December 14, 2015

1. Meiyue Shao 說道：

利用矩阵乘法 $\begin{bmatrix} I & I \\ 0 & I \end{bmatrix} \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix} \begin{bmatrix} -B & I \\ A & I \end{bmatrix} =\begin{bmatrix} 0 & A+B \\ AB & B \end{bmatrix}$ 可以得到另一个证明

• ccjou 說道：

謝謝，補充了矩陣乘法的證明。