Linear Algebra Exercises-n-Answers.pdf

208 Linear Algebra, by Hefferon
Five.IV.2.26
(a) The characteristic polynomial is c(x) = x(x − 1). For λ 1 = 0 we have
( ) −y ∣∣
N (t − 0) = { y ∈ C}
y
(of course, the null space of t 2 is the same). For λ 2 = 1,
( )
x ∣∣
N (t − 1) = { x ∈ C}
0
(and the null space of (t − 1) 2 is the same). We can take this basis
( ) (
1 1
B = 〈 , 〉
−1 0)
to get the diagonalization.
( ) −1 ( ) ( )
1 1 1 1 1 1
=
−1 0 0 0 −1 0
( )
0 0
0 1
(b) The characteristic polynomial is c(x) = x 2 − 1 = (x + 1)(x − 1). For λ 1 = −1,
( )
−y ∣∣
N (t + 1) = { y ∈ C}
y
and the null space of (t + 1) 2 is the same. For λ 2 = 1
( ) y ∣∣
N (t − 1) = { y ∈ C}
y
and the null space of (t − 1) 2 is the same. We can take this basis
( ) (
1 1
B = 〈 , 〉
−1 1)
to get a diagonalization.
( ) −1 ( ) ( )
1 1 0 1 1 1
=
1 −1 1 0 −1 1
( ) −1 0
0 1
Five.IV.2.27 The transformation d/dx: P 3 → P 3 is nilpotent. Its action on B = 〈x 3 , 3x 2 , 6x, 6〉 is
x 3 ↦→ 3x 2 ↦→ 6x ↦→ 6 ↦→ 0. Its Jordan form is its canonical form as a nilpotent matrix.
⎛ ⎞
0 0 0 0
J = ⎜1 0 0 0
⎟
⎝0 1 0 0⎠
0 0 1 0
Five.IV.2.28 Yes. Each has the characteristic polynomial (x + 1) 2 . Calculations of the powers of
T 1 + 1 · I and T 2 + 1 · I gives these two.
( ) ( )
y/2 ∣∣ 0 ∣∣
N (t 1 + 1) = { y ∈ C} N (t
y
2 + 1) = { y ∈ C}
y
(Of course, for each the null space of the square is the entire space.) The way that the nullities rise
shows that each is similar to this Jordan form
(
matrix
)
−1 0
1 −1
and they are therefore similar to each other.
Five.IV.2.29 Its characteristic polynomial is c(x) = x 2 + 1 which has complex roots x 2 + 1 = (x +
i)(x − i). Because the roots are distinct, the matrix is diagonalizable and its Jordan form is that
diagonal matrix. (
−i
)
0
0 i
To find an associated basis we compute the null spaces.
( ) ( )
−iy ∣∣ iy ∣∣
N (t + i) = { y ∈ C} N (t − i) = { y ∈ C}
y
y
For instance,
T + i · I =
( )
i −1
1 i