Proof of $\text{rank}(A+B)\leq \text{rank}(A)+\text{rank}(B)$ by another way.

Here's the problem:

Let $A,B\in Mat_{n}(\mathbb R)$. Use $ \begin{pmatrix} A & A \\ A & A+B \\ \end{pmatrix} $ to show that $$\text{rank}(A+B)\le \text{rank}(A)+\text{rank}(B)$$

By performing elementary operations, I've got that $\text{rank}\begin{pmatrix} A & A \\ A & A+B \\ \end{pmatrix}=\text{rank}\begin{pmatrix} A & O \\ O & B \\ \end{pmatrix}=\text{rank}(A)+\text{rank}(B)$. My intuitive approach is say that since $\begin{pmatrix} A & A \\ A & A+B \\ \end{pmatrix}$ contains $\begin{pmatrix}A+B \\ \end{pmatrix}$, $$\text{rank}(A)+\text{rank}(B)=\text{rank}\begin{pmatrix} A & A \\ A & A+B \\ \end{pmatrix}\geq \text{rank}\begin{pmatrix}A+B \\ \end{pmatrix}=\text{rank}(A+B)$$ but I can't prove this claim well. Could someone give me some hints? Thanks in advance!

Solution 1:

$$A + B = \begin{bmatrix}I&I\end{bmatrix} \begin{bmatrix} A\\B\end{bmatrix}$$

Therefore, $\operatorname{rank} (A+B) \leq \operatorname{rank} \begin{bmatrix} A\\B\end{bmatrix}$. Since $\operatorname{rank} \begin{bmatrix} A\\B\end{bmatrix} \leq \operatorname{rank} (A) + \operatorname{rank} (B)$, we have the desired inequality.