Graphics Reference
In-Depth Information
Solution.
We need to solve the equation
2
l
I-= -=
l
40
for l. The solutions are l=2 or -2, which are the eigenvalues of T. Next, to find the
eigenvectors, we need to solve
()
=
xyA
l
()
xy
,
that is,
(
)
1
-
l
x
+
y
=
0
(
)
31
x
-+
l
y
=
0
.
When l=2, this leads to x = y. In other words, (1,1) is an eigenvector for the eigen-
value 2 and is a basis for its eigenspace. Similarly, when l=-2, then y =-3x, and
(1,-3) is an eigenvector for the eigenvalue -2 and is a basis for its eigenspace.
We can use Theorem C.4.7 to show that not every linear transformation on a vector
space over the reals is diagonalizable. For example, consider the transformation on
R
2
represented by the matrix
01
10
Ê
Ë
ˆ
¯
.
-
The characteristic polynomial of this transformation is t
2
+ 1, which has no real
root. Of course, over the complex numbers every polynomial has a root, so that linear
transformations over complex vector spaces always have eigenvalues and eigenvectors.
The next proposition lists two properties of the characteristic polynomial which
sometimes come in handy, especially in the two-dimensional case where they com-
pletely characterize the polynomial.
C.4.9. Proposition.
Let A = (a
ij
) be an n ¥ n matrix and let
()
=
n
n
-
1
pt
t
+
a
t
+ ◊◊◊+
a t
+
a
n
-
1
1
0
be its characteristic polynomial. Then
(1) a
n-1
=-tr (A),
(2) a
0
= (-1)
n
det (A), and
(3) if n = 2, then
2
()
=-
()
(
)
+
()
pt
t
trA t
det
A
.
Proof.
Using properties of the determinant it is easy to see that
()
=-
(
)
(
)
◊◊◊
(
)
+
pt
ta ta
-
ta
nn
-
polynomial in t of degree n
-
2
.
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