Given the following block matrix: $$ C=\left[
\begin{array}{c|c}
I&A\\
\hline
0&B
\end{array}
\right] $$
With $A,B,I,0 \in M_{n,n}(K)$ show that the rank of $C$ is $n+rk(B)$
To do this I've shown that the rank of $ \left[\!\!
\begin{array}{c|c}
I&A\\
\end{array}\!\!
\right] $ is equal to $n$ since $I$ is a basis of $K^n$ and thus none of the columns of $A$ can be linearly independent with those of $I$.
Also I've argued that the rank of $ \left[\!\!
\begin{array}{c|c}
0&B\\
\end{array}\!\!
\right] $ is just $\operatorname{rk}(B)$ because the columns of $0$ are exactly $0_{K^n}$ and thus cannot be linearly independent with the columns of $B$.
Of course both arguments are the same if we consider rows instead of columns.
To end my argument I would need to show that: $$\operatorname{rk}(C) = \operatorname{rk}(\left[\!\!
\begin{array}{c|c}
I&A\\
\end{array}\!\!
\right])+\operatorname{rk}(\left[\!\!
\begin{array}{c|c}
0&B\\
\end{array}
\!\!\right])=n+rk(B)$$
Which is quite clear but I'm not sure how to formalize it, any suggestions?
Best Answer
The rank is invariant under elementary row/column operations, that is adding or subtracting a row/column to or from another row/column.
Using elementary column operations you can kill off all of $A$ thanks to the columns of $I$, and these operations do not change $B$.
In the remaining block matrix, you see that the columns on the left are independent from the columns of the right, so that the ranks add up. [Hint on how to prove this properly: think of how you would prove properly that the $2\times 2$ identity matrix has rank $2$].