Linear Algebra: Upper Triangular Matrix 
How do I go about solving the following parts? I've tried writing the inequalities for s≤t and i>j for the given 4 mij but always ended up with i less than j? Am I missing something here or is my approach completely wrong? Thanks for the help!
 A: Note that $$(M_{st})_{ij} = \begin{cases} a_{ij} & i < s, j<t \\a_{(i+1)j} & i \geq s, j<t \\ a_{i(j+1)} & i<s ,j \geq t \\ a_{(i+1)(j+1)} & i\geq s ,j \geq t\end{cases}$$.
Given that $s<t$ and $i > j$, we have to prove that $a_{ij} = 0$. Now, it is not possible that $i \leq s$ and $j  > t$ as $s \geq i > j >t > s$ is a contradiction, so the third case is ruled out. 
If case $1$ happens, then $a_{ij}$ is the element at the index $ij$. Since $i > j$, $(M_{st})_{ij} = a_{ij} = 0$ because $A$ itself is  upper triangular.
If case $2$ happens, then $a_{(i+1)j}$ is the element at the index $ij$. Since $i > j$, $i+1>j$, hence $(M_{st})_{ij} = a_{(i+1)j} = 0$ because $A$ itself is  upper triangular.
If case $4$ happens, then $a_{(i+1)(j+1)}$ is the element at the index $ij$. Since $i > j$, $i+1 > j+1$, hence $(M_{st})_{ij} = a_{(i+1)(j+1)} = 0$ because $A$ itself is  upper triangular.
Hence, whatever happens, $i > j \implies (M_{st})_{ij} = 0$ Hence, $M_{st}$ is an upper triangular matrix.
For the second one, I want you to go through the above analysis again and show that $M_{ss} = 0$ if $s < t$. Hence all the diagonal entries will be $0$, hence the matrix is strictly upper triangular, hence the determinant is zero. 
The answer to the third one I leave to you with a hint: Break the matrix into it's diagonal and non-diagonal parts, call them $D$ and $N$. $N$ is  nilpotent  as $N^n = 0$, hence $I+N$ is invertible,and $D$ is invertible, and both of these  have upper triangular inverses ($D^{-1}$ is diagonal, and $(N+I)^{-1} = I-N+\ldots + (-1)^{n-1}N^{n-1}$). Now,note that $A=D(I+N)$ hence the inverse of $A$ is the product of upper triangular matrices.
