Graphics Reference
In-Depth Information
C.4
Eigenvalues and Eigenvectors
Let V be a vector space over a field k and T : V Æ V a linear transformation.
Definition. A scalar l in k is called an eigenvalue of T if there exists a nonzero vector
v in V such that
() =l .
T vv
(C.3)
Every vector satisfying equation (C.3) is called an eigenvector for the eigenvalue l. The
set of eigenvectors for an eigenvalue l is called the eigenspace of l. If A is an n ¥ n
matrix over k, then the eigenvalues and eigenvectors of A are the eigenvalues and eigen-
vectors, respectively, of the linear transformation T : k n
Æ k n associated to A.
C.4.1. Lemma. The eigenspace of an eigenvalue l is the kernel of the transforma-
tion T -lI and hence is a vector subspace.
Proof.
Straightforward.
Definition. A linear transformation T : V Æ V that can be represented by a diagonal
matrix with respect to some basis of V is said to be diagonalized by the basis, or simply
diagonalizable . An n ¥ n matrix is diagonalizable if the associated linear transforma-
tion on k n is diagonalizable.
C.4.2. Theorem. A linear transformation T : V Æ V is diagonalizable if and only if
V has a basis of eigenvectors of T. The diagonal entries of the matrix with respect to
the basis that diagonalizes the transformation are then the eigenvalues of T.
Proof.
Easy.
C.4.3. Theorem. Let T : V Æ V be a linear transformation. If v 1 , v 2 ,..., v m are
nonzero eigenvectors for T corresponding to distinct eigenvalues l 1 , l 2 ,..., l m ,
respectively, then the vectors v 1 , v 2 ,..., v m are linearly independent.
Proof. The proof is by induction on m. The case m = 1 is clear. Assume that the
theorem has been proved for m - 1, m > 1. Assume that the m eigenvectors v 1 , v 2 ,
..., v m satisfy a relation
0v v
=
a
+
a
+ ◊◊◊+
a mm
v
,
(C.4)
11
22
for some a i . Applying T to both sides gives
()
00
v
=
T
aT
() +
() + ◊◊◊+
(
)
=
aT
v
a T
v
11 22
mm
(C.5)
=
ava v
l
+
l
+ ◊◊◊+
a
l
v
111
222
mmm
On the other hand, multiplying (C.4) by l m and subtracting from (C.5) gives
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