Page 259 - 35Linear Algebra

14.3 Relating Orthonormal Bases 259

T
We would like to calculate the product PP . For that, we first develop a
dirty trick for products of dot products:

T
T
T
T
(u v)(w z) = (u v)(w z) = u (vw )z .
T
The object vw is the square matrix made from the outer product of v and w.
T
Now we are ready to compute the components of the matrix product PP .
X X T T
(u j w i )(w i u k ) = (u w i )(w u k )
j i
i i
" #
X T
T
= u
j (w i w ) u k
i
i
(∗) T
= u I n u k
j
T
= u u k = δ jk .
j
The equality (∗) is explained below. Assuming (∗) holds, we have shown that
T
PP = I n , which implies that
T
P = P −1 .

P T
The equality in the line (∗) says that w i w = I n . To see this, we
i i
P T j
examine w i w v for an arbitrary vector v. We can find constants c
i
j
i P
such that v = c w j , so that
j
! ! !
X T X T X j
w i w v = w i w c w j
i i
i i j
X X
T
= c j w i w w j
i
j i
X j X
= c w i δ ij
j i
X
j
= c w j since all terms with i 6= j vanish
j
= v.
P T
Thus, as a linear transformation, w i w = I n fixes every vector, and thus
i i
must be the identity I n .
259