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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?!

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  • $\begingroup$ See this $\endgroup$ – user376343 Sep 17 '18 at 14:55
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You cannot say the equality is true. For example,

$$\mathrm{rank}\left(\begin{bmatrix}0&0\\1&1\end{bmatrix}\right)\neq\mathrm{rank}\left(\begin{bmatrix}0\\1\end{bmatrix}\right)+\mathrm{rank}\left(\begin{bmatrix}0\\1\end{bmatrix}\right)$$

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  • $\begingroup$ Yes, You are right. But under what conditions we can say that?! $\endgroup$ – im hs Sep 17 '18 at 12:54
  • $\begingroup$ It should be true since I see this in many papers and books! but without any proof and refer! it seems that it should be trivial!! $\endgroup$ – im hs Sep 17 '18 at 12:54
  • $\begingroup$ and how about if $A$ is invertible? $\endgroup$ – im hs Sep 17 '18 at 12:56
  • $\begingroup$ @imhs please refer us to an example of a paper or book saying such a thing. "should be true" says nothing if you are given a counter example. $\endgroup$ – GuySa Sep 17 '18 at 13:30
  • $\begingroup$ @GuySa please look at this for example: robust model-based fault diagnosis for dynamic systems by Chen and Ron Patton on page 75. $\endgroup$ – im hs Sep 17 '18 at 16:57

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