Digital Signal Processing Reference
In-Depth Information
Similarity transformations have been demonstrated through examples. Certain im-
portant properties of these transformations are derived next. Consider first the
determinant of
(z
I
-
A
v
).
From (13.52),
det(z
I
-
A
v
) = det(z
I
-
P
-1
AP
) = det(z
P
-1
IP
-
P
-1
AP
)
= det[
P
-1
(z
I
-
A
)
P
].
(13.53)
For two square matrices,
det
R
1
R
2
= det
R
1
det
R
2
.
(13.54)
Then we can express (13.53) as
det(z
I
-
A
v
) = det
P
-1
det(z
I
-
A
) det
P
,
(13.55)
R, R
-1
R
=
I
.
because, for a matrix
Thus,
det
R
-1
R
= det
R
-1
det
R
= det
I
= 1.
(13.56)
Hence, (13.55) yields the first property:
det(z
I
-
A
v
) = det(z
I
-
A
).
(13.57)
The roots of are the
characteristic values
, or the
eigenvalues
, of
A
.
(See Appendix G.) Thus, the eigenvalues of are equal to those of
A
, from
(13.57). Since the transfer function is unchanged under a similarity transformation,
and since the eigenvalues are the poles of the transfer function, we are not surprised
that they are unchanged.
A second property is now derived. From (13.57) with
det(z
I
-
A
)
A
v
z = 0,
det
A
v
= det
A
.
(13.58)
The determinant of is equal to the determinant of
A
. This property can also be
seen from the fact that the determinant of a matrix is equal to the product of its
eigenvalues. (See Appendix G.)
The third property of a similarity transformation can also be seen from the
fact that the eigenvalues of and of
A
are equal. The trace (sum of the diagonal
elements) of a matrix is equal to the sum of the eigenvalues; hence,
A
v
A
v
tr
A
v
= tr
A
.
(13.59)
A fourth property was demonstrated in Example 13.12. Since the transfer
function is unchanged under a similarity transformation,
C
v
(z
I
-
A
v
)
-1
B
v
+ D
v
=
C
(z
I
-
A
)
-1
B
+ D.
(13.60)
The proof of this property is left as an exercise.
