By choosing an orthogonal basis
q
j
,
k
1≤k≤nj
q
j
,
k
1
k
nj
for each
RPj
Pj
and collecting the basis vectors in
Qj=(
q
j
,
1
q
j
,
2
…
q
j
,
nj
)
Qj
q
j
,
1
q
j
,
2
…
q
j
,
nj
we find that
Pj=QjQjT=∑k=1nj
q
j
,
k
q
j
,
k
T
Pj
Qj
Qj
k
1
nj
q
j
,
k
q
j
,
k
As a result, the spectral
representation takes the form

B=∑j=1hλjQjQjT=∑j=1hλj∑k=1nj
q
j
,
k
q
j
,
k
T
B
j
1
h
λj
Qj
Qj
j
1
h
λj
k
1
nj
q
j
,
k
q
j
,
k

(1)
This is the spectral representation in perhaps its most detailed
dress. There exists, however, still another form! It is a form
that you are likely to see in future engineering courses and is
achieved by assembling the

QjQj
into a single

nn-by-

nn
orthonormal matrix

Q=(
Q1…Qh
)
Q
Q1
…
Qh
Having orthonormal columns it follows that

QTQ=I
Q
Q
I
.

QQ being square, it
follows in addition that

QT=Q-1
Q
Q
.
Now

B
q
j
,
k
=λj
q
j
,
k
B
q
j
,
k
λj
q
j
,
k
may be encoded in matrix terms via

where

ΛΛ is the

nn-by-

nn diagonal matrix whose first

n1n1
diagonal terms are

λ1λ1,
whose next

n2n2
diagonal terms are

λ2λ2,
and so on. That is, each

λjλj
is repeated according to its multiplicity. Multiplying each
side of

Equation 2, from the right, by

QT
Q
we arrive at

Because one may just as easily write

one says that

QQ
diagonalizes BB.

Let us return the our example
B=(
111
111
111
)
B
1
1
1
1
1
1
1
1
1
of the last chapter. Recall that the eigenspace associated with
λ1=0
λ1
0
had
e
1
,
1
=(
-1
1
0
)
e
1
,
1
-1
1
0
and
e
1
,
2
=(
-1
0
1
)
e
1
,
2
-1
0
1
for a basis. Via Gram-Schmidt we may replace this with
q
1
,
1
=12(
-1
1
0
)
q
1
,
1
1
2
-1
1
0
and
q
1
,
2
=16(
-1
-1
2
)
q
1
,
2
1
6
-1
-1
2
Normalizing the vector associated with
λ2=3
λ2
3
we arrive at
q
2
,
1
=13(
1
1
1
)
q
2
,
1
1
3
1
1
1
and hence
Q=(
q11q12q2
)=16(
−3-12
3-12
022
)
Q
q11
q12
q2
1
6
3
-1
2
3
-1
2
0
2
2
and
Λ=(
000
000
003
)
Λ
0
0
0
0
0
0
0
0
3