linear-guest

14.3 Relating Orthonormal Bases 259
We would like to calculate the product P P T . For that, we first develop a
dirty trick for products of dot products:
(u v)(w z) = (u T v)(w T z) = u T (vw T )z .
The object vw T is the square matrix made from the outer product of v and w.
Now we are ready to compute the components of the matrix product P P T .
∑
(u j w i )(w i u k ) = ∑ (u T j w i )(wi T u k )
i
i
[ ]
∑
= u T j (w i wi T ) u k
i
(∗)
= u T j I n u k
= u T j u k = δ jk .
The equality (∗) is explained below. Assuming (∗) holds, we have shown that
P P T = I n , which implies that
P T = P −1 .
The equality in the line (∗) says that ∑ i w iwi
T = I n . To see this, we
examine (∑ i w )
iwi
T v for an arbitrary vector v. We can find constants c
j
such that v = ∑ j cj w j , so that
( ∑
i
w i w T i
)
v =
( ∑
i
w i w T i
) ( ∑
j
c j w j
)
= ∑ j
c j ∑ i
w i w T i w j
= ∑ j
c j ∑ i
w i δ ij
= ∑ j
c j w j since all terms with i ≠ j vanish
= v.
Thus, as a linear transformation, ∑ i w iwi
T
must be the identity I n .
= I n fixes every vector, and thus
259