Linear Algebra

356 Chapter 5. Similarity
That is, one solution to (∗) isλ1 = 3, and we have a first basis vector.
�β1 =
� �
1
0
In the x = 1 possibility, the first equation in (∗) is2· b1 +2· b2 =0,andso
associated with 1 are vectors whose second component is the negative of their
first component.
� �� � � �
3 2 b1 b1
=1·
0 1 −b1 −b1
Thus, another solution is λ2 = 1 and a second basis vector is this.
� �
�β2
1
=
−1
To finish, drawing the similarity diagram
t
−−−−→
T
R2 w.r.t. R E2
2 w.r.t. E2
⏐
⏐
⏐
⏐
id�
id�
R 2 w.r.t. B
t
−−−−→
D
R 2 w.r.t. B
and noting that the matrix Rep B,E2 (id) is easy leads to this diagonalization.
� � � �−1 � �� �
3 0 1 1 3 2 1 1
=
0 1 0 −1 0 1 0 −1
In the next subsection, we will expand on that example by considering more
closely the property of Corollary 2.4. This includes seeing another way, the way
that we will routinely use, to find the λ’s.
Exercises
� 2.6 Repeat Example 2.5 for the matrix from Example 2.2.
2.7 Diagonalize � �these
upper � triangular � matrices.
−2 1
5 4
(a)
(b)
0 2
0 1
� 2.8 What form do the powers of a diagonal matrix have?
2.9 Give two same-sized diagonal matrices that are not similar. Must any two
different diagonal matrices come from different similarity classes?
2.10 Give a nonsingular diagonal matrix. Can a diagonal matrix ever be singular?
� 2.11 Show that the inverse of a diagonal matrix is the diagonal of the the inverses,
if no element on that diagonal is zero. What happens when a diagonal entry is
zero?