1.4.5

Let $$P$$ by a permutation matrix. We will show that $$PP^\text{T} = I$$, which implies that $$P^\text{T} = P^{-1}.$$ Let $$C = PP^\text{T}$$. Then $$c_{ii} = \sum p_{ii}^2 = 1,$$ and $$i$$, $$j$$ s.t. $$i \not= j$$, $$c_{ij} = (\text{row } i)^\text{T} (\text{row } j) = 0$$ because there is at most one 1 per column in $$P$$ (so all terms in the sum will be 0 – this is more obvious graphically). This implies that $$C = I.$$