Prove that a submatrix of an upper triangular matrix is upper triangular I can't see the intuition behind this problem - any help is appreciated. Much thanks! 
Let $ \mathbf{B} $ be the submatrix obtained from upper triangular matrix $ \mathbf{A} $ by deleting the $m$-th row and the $n$-th column, prove that $ \mathbf{B} $ is upper triangular if $ m \leq n $. 
 A: Let $B$ the resulting matrix with entries $b_{i,j}$. 
The set of $(i,j)$ indices can be partitioned into the following (disjoint) 4 four cases:
$$ \underbrace{(i<m \ and \ j<n)}_\text{[1]} \ \ or \ \
\underbrace{(i\geq m \ and \ j <n)}_\text{[2]} \ \ or \ \
\underbrace{(i<m \ and \ j \geq n)}_\text{[3]} \ \ or \ \ 
\underbrace{(i\geq m \ and \ j \geq n)}_\text{[4]}$$
with
$$\tag{1}b_{i,j}=\begin{cases}
a_{i,j} &\text{in case [1]}\\
a_{i+1,j} &\text{in case [2]}\\
a_{i,j+1} &\text{in case [3]}\\
a_{i+1,j+1} &\text{in case [4]}
 \end{cases}$$
In order to prove that $B$ is upper triangular, we have to prove that 
$$\tag{2}i>j \Longrightarrow b_{i,j}=0$$
knowing that the same is true for $A$:
$$\tag{3}i>j \Longrightarrow a_{i,j}=0$$
(2) will be a consequence of the separate consideration of the four cases:
$$\tag{1}\text{If} \  i>j, \ \ b_{i,j}=\begin{cases}
[1] & a_{i,j}=0 &\text{due directly to (3)}&\\
[2] & a_{i+1,j}=0 &\text{due to (3) because} & i>j \Rightarrow i+1>i>j\\
[3] & a_{i,j+1}=0 &\text{see remark below.}& \\
[4] & a_{i+1,j+1}=0 &\text{due to (3) because}  & i>j \Rightarrow i+1>j+1\\
 \end{cases}$$
Remark : In fact, under the hypotheses taken, case [3] cannot occur because, referring to its defining inequalities ($i<m$ and $j\geq n$), we would have:
$$m > i > j \geq n \ \ \ \text{contradicting assumption} \  \ \ n>m.$$
A: Upper triangular means that $A_{kj}=0$ whenever $k>j$. We should check the same for $B$. 
So consider $B_{kj}$, with $k>j$. We have the following two cases:


*

*$k<m$. In this case, $j<k<m\leq n$, so $B_{kj}=A_{kj}=0$.

*$k\geq m$, $j<n$. In this case, $B_{kj}=A_{k-1,j}$. If $k-1>j$, then $A_{k-1,j}=0$; otherwise, $k-1\leq j$ and $k>j$, so $k=j+1$. In this last case, $B_{kj}=A_{j+1,j}=0$. 
So $B$ is upper triangular. 
