# Rank of block triangular matrix

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?!

• See this – user376343 Sep 17 '18 at 14:55

$$\mathrm{rank}\left(\begin{bmatrix}0&0\\1&1\end{bmatrix}\right)\neq\mathrm{rank}\left(\begin{bmatrix}0\\1\end{bmatrix}\right)+\mathrm{rank}\left(\begin{bmatrix}0\\1\end{bmatrix}\right)$$
• and how about if $A$ is invertible? – im hs Sep 17 '18 at 12:56