Let $A$ be an upper triangular matrix and $B_{ij}$ be the matrix that results when the $i$th row and $j$th column of $A$ are deleted.
Why is it that when $i
I am not clear as to how this property will always hold for any $n\times n$ matrix.
Let $A$ be an upper triangular matrix and $B_{ij}$ be the matrix that results when the $i$th row and $j$th column of $A$ are deleted.
Why is it that when $i
I am not clear as to how this property will always hold for any $n\times n$ matrix.
A matrix, $A=(a_{ij})$ is upper triangular if $a_{ij}=0$ when $i>j$ (further down rows than over columns). Assume $k<\ell$ and consider $B_{k\ell}=(b_{ij})$ (the matrix obtained by deleting the $k^{th}$-row and $\ell^{th}$-column of $A$).
We need to show that $b_{ij}=0$ for $i>j$.
Case: $i
Case: $i \geq k$ and $j<\ell$ then $b_{ij}=a_{i+1,j}=0$ since $i+1>i>j$.
Case: $i \geq k$ and $j \geq \ell$ then $b_{ij}=a_{i+1,j+1}=0$ since $i+1>j+1$ because $i>j$.
Case: $i
Thus $b_{ij}=0$ for all $i>j$ and so $B_{k\ell}$ is upper-triangular.