Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Digital systems, linear algebra, introduction to formal proof / 5 / b

The permutation matrix isRows three and four of the identity matrix have been interchanged, so left multiplication interchanges those rows of any matrix.

Left multiplication by $E$ interchanges rows three and four; applying it twice restores every row. Equivalently, it sends the standard basis vectors $e_3,e_4$ to each other and fixes $e_1,e_2$, so its square fixes a basis. $E^2=I$.

Since $E^2=I$, the same matrix is both a left and a right inverse. $E^{-1}=E$.

The $j$th column of $AE$ is $A(E{e}_j)$. The permutation matrix fixes $e_1,e_2$ and interchanges $e_3,e_4$, so these columns are $A{e}_1,A{e}_2,A{e}_4,A{e}_3$. They are exactly the first, second, fourth and third columns of $A$. Thus right multiplication by $E$ interchanges columns three and four, leaving the others unchanged.

The first left multiplication swaps rows three and four. The final right multiplication swaps columns three and four, restoring the two diagonal ones to their diagonal positions while exchanging the earlier entries of the two rows. ThereforeThe two specified zeros at positions $(3,4)$ and $(4,3)$ are important. Conjugation by this permutation matrix preserves this restricted lower triangular pattern; it would not preserve lower triangularity for an arbitrary matrix with a nonzero $(4,3)$ entry.