Transpose operator
AT
A
flips the matrix across it's diagonal.
A=(
a1,1a1,2
a2,1a2,2
)
A
a
1
1
a
1
2
a
2
1
a
2
2
AT=(
a1,1a2,1
a1,2a2,2
)
A
a
1
1
a
2
1
a
1
2
a
2
2
Column
i
i of
A
A is row
i
i of
AT
A
Recall, inner product
x=(
x
0
x
1
⋮
x
n

1
)
x
x
0
x
1
⋮
x
n

1
y=(
y
0
y
1
⋮
y
n

1
)
y
y
0
y
1
⋮
y
n

1
xTy=(
x
0
x
1
…
x
n

1
)(
y
0
y
1
⋮
y
n

1
)=∑
i
xiyi=〈y,x〉
x
y
x
0
x
1
…
x
n

1
y
0
y
1
⋮
y
n

1
i
x
i
y
i
y
x
on
Rn
n
Hermitian transpose
AH
A
, transpose and conjugate
AH=AT*
A
A
〈y,x〉=xHy=∑
i
xiyi*
y
x
x
y
i
x
i
y
i
on
Cn
n
Now, let
b
0
b
1
…
b
n

1
b
0
b
1
…
b
n

1
be an orthonormal basis for
Cn
n
i=01…n−1〈
b
i
,
b
i
〉=1 ,
i
i
0
1
…
n
1
b
i
b
i
1
i≠j〈
b
i
,
b
j
〉=
b
j
H
b
i
=0
i
j
b
i
b
j
b
j
b
i
0
Basis matrix:
B=(
⋮⋮⋮
b
0
b
1
…
b
n

1
⋮⋮⋮
)
B
⋮
⋮
⋮
b
0
b
1
…
b
n

1
⋮
⋮
⋮
Now,
BHB=(
…
b
0
H…
…
b
1
H…
⋮
…
b
n

1
H…
)(
⋮⋮⋮
b
0
b
1
…
b
n

1
⋮⋮⋮
)=(
b
0
H
b
0
b
0
H
b
1
…
b
0
H
b
n

1
b
1
H
b
0
b
1
H
b
1
…
b
1
H
b
n

1
⋮
b
n

1
H
b
0
b
n

1
H
b
1
…
b
n

1
H
b
n

1
)
B
B
…
b
0
…
…
b
1
…
⋮
…
b
n

1
…
⋮
⋮
⋮
b
0
b
1
…
b
n

1
⋮
⋮
⋮
b
0
b
0
b
0
b
1
…
b
0
b
n

1
b
1
b
0
b
1
b
1
…
b
1
b
n

1
⋮
b
n

1
b
0
b
n

1
b
1
…
b
n

1
b
n

1
For orthonormal basis with basis matrix
B
B
BH=B1
B
B
(
BT=B1
B
B
in
Rn
n
)
BH
B
is easy to calculate while
B1
B
is hard to calculate.
So, to find
α
0
α
1
…
α
n

1
α
0
α
1
…
α
n

1
such that
x=∑
i
α
i
b
i
x
i
α
i
b
i
Calculate
(α=B1x)⇒(α=BHx)
α
B
x
α
B
x
Using an orthonormal basis we rid ourselves of the inverse operation.
"My introduction to signal processing course at Rice University."