Prove that $$\operatorname{rank}\begin{bmatrix} A & B \\ 0 & C \end{bmatrix}\geq \operatorname{rank}A+\operatorname{rank}C.$$
Deduce that the rank of an upper triangular matrix is not less than
the number of non-zero diagonal elements.
If I take the column bases of $A$ and $C$, the corresponding columns of the bigger matrix are Linearly Independent.
This above idea is fairly intuitive and very helpful, but I am unable to write it down as a formal proof. Could someone help me express it in a clean and presentable form?
For the deduction part, I see that any Upper Triangular Matrix can be expressed as $ \begin{bmatrix} A & B \\ 0 & C \\ \end{bmatrix}$.
Now, if I could show that:
rank of $A \geq$ number of non-zero diagonal elements in A $\ \ \ \ \ \ \ \ -(1)$
rank of $C \geq$ number of non-zero diagonal elements in C $\ \ \ \ \ \ \ \ -(2)$
then the deduction logically follows.
The thing I find strange is that $A$ and $C$ are both themselves Upper Triangular Matrices. So effectively, both $(1)$ and $(2)$ are exactly what I started out to deduce in the first place. This basically puts me in a loop. I seem to be missing something and would like some help here. Thanks.
Best Answer
Hint. Let $r,s$ be the ranks of $A$ and $C$ respectively, $\{Ax_1,\ldots,Ax_r\}$ be a basis of the column space of $A$ and $\{Cy_1,\ldots,Cy_s\}$ be a basis of the column space of $C$. Now show that the vectors $$ \pmatrix{Ax_1\\ 0},\ldots,\pmatrix{Ax_r\\ 0},\pmatrix{By_1\\ Cy_1},\ldots,\pmatrix{By_s\\ Cy_s} $$ are linearly independent and they all reside in the image of $\pmatrix{A&B\\ 0&C}$.