Linear Algebra

Section II. Similarity 357
2.12 The equation ending Example 2.5
� �−1 � ��
1 1 3 2 1
0 −1 0 1 0
� �
1 3
=
−1 0
�
0
1
is a bit jarring because for P we must take the first matrix, which is shown as an
inverse, and for P −1 we take the inverse of the first matrix, so that the two −1
powers cancel and this matrix is shown without a superscript −1.
(a) Check that this nicer-appearing equation holds.
� � � �� ��
3 0 1 1 3 2 1
=
0 1 0 −1 0 1 0
�−1 1
−1
(b) Is the previous item a coincidence? Or can we always switch the P and the
P −1 ?
2.13 Show that the P used to diagonalize in Example 2.5 is not unique.
2.14 Find a formula for the powers of this matrix Hint: see Exercise 8.
� �
−3 1
−4 2
� 2.15 Diagonalize � � these. � �
1 1
0 1
(a)
(b)
0 0
1 0
2.16 We can ask how diagonalization interacts with the matrix operations. Assume
that t, s: V → V are each diagonalizable. Is ct diagonalizable for all scalars c?
What about t + s? t ◦ s?
� 2.17 Show that matrices of this form are not diagonalizable.
� �
1 c
c �= 0
0 1
2.18 Show � that � each of these � is� diagonalizable.
1 2
x y
(a)
(b)
x, y, z scalars
2 1
y z
5.II.3 Eigenvalues and Eigenvectors
In this subsection we will focus on the property of Corollary 2.4.
3.1 Definition A transformation t: V → V has a scalar eigenvalue λ if there
is a nonzero eigenvector � ζ ∈ V such that t( � ζ)=λ · � ζ.
(“Eigen” is German for “characteristic of” or “peculiar to”; some authors call
these characteristic values and vectors. No authors call them “peculiar”.)
3.2 Example The projection map
⎛
⎝ x
⎞
y⎠
z
π
⎛
↦−→ ⎝ x
⎞
y⎠
x, y, z ∈ C
0