Image Processing Reference
In-Depth Information
first-order tensor scalar products are therefore the same as those of the scalars and
the vectors of linear algebra. Not surprisingly, the second-order tensors can also be
equipped with a scalar product. We restate these results for the first- and second-order
tensors.
Lemma 3.3.
With the following scalar products
=
k
u
(
k
)
∗
v
(
k
)
u
,
v
(3.44)
=
kl
A
(
k, l
)
∗
B
(
kl
)
A
,
B
(3.45)
where
k, l
run over the components of the tensor indices, the tensors up to the second-
order having the same dimension constitute Hilbert spaces.
Example 3.4. Weattempttoexpand3
×
3 imagesinanorthogonalbasiswithmatrix
elements having
1 or 0, in an analogous manner as in Example 3.3. We note, how-
ever, that it is now more difficult to guess orthogonal matrices. We suggest “guess-
ing” but only three variables instead of 9, by using the following observation.
Consider the vectors
u
(
i
) in
E
3
:
±
u
0
=( 0
,
1
,
0)
T
u
1
=(
−
1
,
0
,
1)
T
u
2
=( 1
,
0
,
1)
T
whichareorthogonalinthesenseoftheordinaryscalarproductoffirst-ordertensors,
i.e.,
=
u
i
u
j
=0
u
i
,
u
j
for
i
=
j
(3.46)
By using the following product based on the ordinary matrix product rule in linear
algebra, tensor product, we can construct 3
3 matrices
U
k
that are orthogonal
second-order tensors in the sense of the above scalar product (Eq. 3.45).
×
U
k
=
u
i
u
j
(3.47)
That the matrices
U
k
are orthogonal to each other follows by direct examination:
u
i
u
j
,
u
i
u
j
=
m
u
i
(
m
)
u
j
(
n
)
u
i
(
m
)
u
j
(
n
)
(3.48)
n
=(
m
u
i
(
m
)
u
i
(
m
))(
n
u
j
(
n
)
u
j
(
n
))
(3.49)
=0
(3.50)
when (
i, j
)
=(
i
,j
). Consequently, we can expand all 3
×
3 real matrices by using
the scalar product in Eq. (3.45) and the basis
U
k
.
