Editing Permutation matrix (section)

== The transpose is also the inverse ==
A related argument proves that, as we claimed above, the transpose of any permutation matrix ''P'' also acts as its inverse, which implies that ''P'' is invertible. (Artin leaves that proof as an exercise,<ref name="Artin Algebra" />{{rp|page=26}} which we here solve.)  If <math>P=(p_{i,j})</math>, then the <math>(i,j)^\text{th}</math> entry of its transpose <math>P^\mathsf{T}</math> is <math>p_{j,i}</math>.  The <math>(i,j)^\text{th}</math> entry of the product <math>PP^\mathsf{T}</math> is then
:<math display=block>\sum_{k=1}^n p_{i,k}p_{j,k}.</math>
Whenever <math>i\ne j</math>, the <math>k^\text{th}</math> term in this sum is the product of two different entries in the <math>k^\text{th}</math> column of ''P''; so all terms are 0, and the sum is 0.  When <math>i=j</math>, we are summing the squares of the entries in the <math>i^\text{th}</math> row of ''P'', so the sum is 1.  The product <math>PP^\mathsf{T}</math> is thus the identity matrix.  A symmetric argument shows the same for <math>P^\mathsf{T}P</math>, implying that ''P'' is invertible with <math>P^{-1}=P^\mathsf{T}</math>.