On sait que : $$\ds\forall j\in\llbracket1,n\rrbracket,\; \vv{e_j'}=\sum_{i=1}^{n}{\left\langle \vv{e_j'},\vv{e_i}\right\rangle \vv{e_i}}\qquad\text{et}\qquad\forall k\in\llbracket1,n\rrbracket,\; \vv{e_k}=\sum_{j=1}^{n}{\left\langle \vv{e_k},\vv{e_j'}\right\rangle \vv{e_j'}}$$ ce qui permet de dire que : $$\ds P=\left(\left\langle \vv{e_j'},\vv{e_i}\right\rangle \right)_{(i,j)\in\llbracket1,n\rrbracket^{2}}\qquad\text{et}\qquad P^{-1}=\left(\left\langle \vv{e_k},\vv{e_j'}\right\rangle \right)_{(j,k)\in\llbracket1,n\rrbracket^{2}}$$ d'où l'on déduit que : $$\ds {}^t\! P=\left(\left\langle \vv{e_j'},\vv{e_i}\right\rangle \right)_{(j,i)\in\llbracket1,n\rrbracket^{2}}=\left(\left\langle \vv{e_i},\vv{e_j'}\right\rangle \right)_{(j,i)\in\llbracket1,n\rrbracket^{2}}=P^{-1}$$