suppose that $C$ is full column rank matrix, can we say that the following equality is true:
$$\operatorname{rank}\biggl(\begin{bmatrix} A & 0\\B & C \end{bmatrix}\biggr)= \operatorname{rank}\biggl(\begin{bmatrix} A \\B \end{bmatrix}\biggr) +\operatorname{rank}(C)$$
I have an idea about it:
$$\begin{bmatrix} A & 0\\B & C \end{bmatrix}=\begin{bmatrix} A & 0\\B & 0 \end{bmatrix}+\begin{bmatrix} 0 & 0\\0 & C \end{bmatrix}$$.
Now using the inequality, $\text{rank}(A+B)\leq \text{rank}(A)+\text{rank}(B)$ results in
$$\operatorname{rank}\biggl(\begin{bmatrix} A & 0\\B & C \end{bmatrix}\biggr)\leq \operatorname{rank}\biggl(\begin{bmatrix} A \\B \end{bmatrix}\biggr) +\operatorname{rank}(C).$$
The equality holds whenever ${R}\left(\begin{bmatrix} A \\B \end{bmatrix}\right) \cap R(C)=0$ and $C\left(\begin{bmatrix} A \\B \end{bmatrix}\right) \cap C(C)=0$.
But how to conclude it in this case?!
You cannot say the equality is true. For example,
$$\mathrm{rank}\left(\begin{bmatrix}0&0\\1&1\end{bmatrix}\right)\neq\mathrm{rank}\left(\begin{bmatrix}0\\1\end{bmatrix}\right)+\mathrm{rank}\left(\begin{bmatrix}0\\1\end{bmatrix}\right)$$