Linear Algebra

358 Chapter 5. Similarity
has an eigenvalue of 1 associated with any eigenvector of the form
⎛
⎝ x
⎞
y⎠
0
where x and y are non-0 scalars. On the other hand, 2 is not an eigenvalue of
π since no non-�0 vector is doubled.
That example shows why the ‘non-�0’ appears in the definition. Disallowing
�0 as an eigenvector eliminates trivial eigenvalues.
3.3 Example The only transformation on the trivial space {�0 } is �0 ↦→ �0. This
map has no eigenvalues because there are no non-�0 vectors �v mapped to a scalar
multiple λ · �v of themselves.
3.4 Example Consider the homomorphism t: P1 →P1 given by c0 + c1x ↦→
(c0 + c1)+(c0 + c1)x. The range of t is one-dimensional. Thus an application of
t to a vector in the range will simply rescale that vector: c + cx ↦→ (2c)+(2c)x.
That is, t has an eigenvalue of 2 associated with eigenvectors of the form c + cx
where c �= 0.
This map also has an eigenvalue of 0 associated with eigenvectors of the form
c − cx where c �= 0.
3.5 Definition A square matrix T has a scalar eigenvalue λ associated with
the non-�0 eigenvector � ζ if T � ζ = λ · � ζ.
3.6 Remark Although this extension from maps to matrices is obvious, there
is a point that must be made. Eigenvalues of a map are also the eigenvalues of
matrices representing that map, and so similar matrices have the same eigenvalues.
But the eigenvectors are different — similar matrices need not have the
same eigenvectors.
For instance, consider again the transformation t: P1 →P1 given by c0 +
c1x ↦→ (c0+c1)+(c0+c1)x. It has an eigenvalue of 2 associated with eigenvectors
of the form c + cx where c �= 0. If we represent t with respect to B = 〈1 +
1x, 1 − 1x〉
� �
2 0
T =RepB,B(t) =
0 0
then 2 is an eigenvalue of T , associated with these eigenvectors.
� � � �� � � � � �
c0 �� 2 0 c0 2c0 c0 ��
{
= } = { c0 ∈ C, c0�= 0}
c1 0 0 c1 2c1 0
On the other hand, representing t with respect to D = 〈2+1x, 1+0x〉 gives
� �
3 0
S =RepD,D(t) =
−3 2