Computing $PAQ = LU$ using Gaussian elimination with complete pivoting > Suppose $PAQ = LU$ is computed via Gaussian elimination with complete pivoting. Show that there is no element in $e_i^{T}U$ i.e., row $i$ of $U$, whose magnitude is larger than $|\mu_{ii}| = |e_i^{T}U e_i|$, i.e., the magnitude of the $(i,i)$ diagonal element of $U$. I understand Gaussian elimination with complete pivoting but I am not really sure what this question is really asking of me to do. Any suggestions or comments is greatly appreciated.

By definition, the absolute value of the $i$th pivot $\mu_{ii}$ maximizes the absolute value of _all_ elements in trailing submatrix. In particular, the $i$th pivot is larger than elements in the first row of the trailing submatrix. This row constitutes the upper triangular elements of the $i$th row of the matrix $U$. In short, for each row of $U$ the element with the largest absolute value is found on the on diagonal.