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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 $i<j$, $B_{ij}$ is upper triangular?

I am not clear as to how this property will always hold for any $n\times n$ matrix.

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Have you actually tried it on an example? If you take a 7 by 7 upper triangular matrix, say, and delete row 2 and column 5, what do you get? The general principle should be hopefully become obvious then. – Hans Lundmark Nov 3 '11 at 7:36
up vote 1 down vote accepted

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: $i<k$ and $j<\ell$ then $b_{ij}=a_{ij}=0$ since $i>j$.

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: $i<k$ and $j \geq \ell$ is impossible since this would imply $i<k<\ell\leq j$ contradicting the fact that $j<i$.

Thus $b_{ij}=0$ for all $i>j$ and so $B_{k\ell}$ is upper-triangular.

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Thank you, your explanation makes it clear. – Sara Nov 11 '11 at 18:33

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