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

Question

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!

Answer 1

Your intuitive approach is correct. There are a few approaches that we could apply to get your inequality; here's one approach. In general, it holds that $\operatorname{rank}(PQ) \leq \min\{\operatorname{rank}(P),\operatorname{rank}(Q)\}$. Thus, if we take $$ P = \pmatrix{A & A\\A&A+B}, \quad Q = \pmatrix{0 & I} $$ (where $I$ denotes the identity matrix), then we have $$ \operatorname{rank}(P)\geq \operatorname{rank}(QP) \geq \operatorname{rank}(QPQ^T) = \operatorname{rank}(A + B). $$

Answer 2

Or like this. For any matrix $A$ $$ \operatorname{rank}\begin{pmatrix} A & *\\* & *\end{pmatrix}\geq \operatorname{rank}(A). $$ An outline of the proof. Choose a nonzero minor $M$ of order $r=\operatorname{rank}(A)$ in the matrix $A$. Then $M$ is a nonzero minor in the large matrix as well.

Answer 3

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

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.