By choosing an orthogonal basis
{
q
j
,
k
|1≤k≤nj}
q
j
,
k
1
k
nj
for each
ℝ
Pj
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
BQ=QΛ
B
Q
Q
Λ
(2)
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
B=QΛQT
B
Q
Λ
Q
(3)
Because one may just as easily write
QTBQ=Λ
Q
B
Q
Λ
(4)
one says that
QQ
diagonalizes BB.
Let us return the our example
B=111111111
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
=-110
e
1
,
1
-1
1
0
and
e
1
,
2
=-101
e
1
,
2
-1
0
1
for a basis. Via Gram-Schmidt we may replace this with
q
1
,
1
=12-110
q
1
,
1
1
2
-1
1
0
and
q
1
,
2
=16-1-12
q
1
,
2
1
6
-1
-1
2
Normalizing the vector associated with
λ2=3
λ2
3
we arrive at
q
2
,
1
=13111
q
2
,
1
1
3
1
1
1
and hence
Q=q11q12q2=16-3-123-12022
Q
q11
q12
q2
1
6
3
-1
2
3
-1
2
0
2
2
and
Λ=000000003
Λ
0
0
0
0
0
0
0
0
3
