Linear Algebra Exercises-n-Answers.pdf

186 Linear Algebra, by Hefferon
(as above, if x, y, z ∈ R then this discriminant is positive so a symmetric, real, 2×2 matrix is similar
to a real diagonal matrix).
For a check we try x = 1, y = 2, z = 1.
b = 0 ± √ 0 + 16
= ∓1 d = 0 ± √ 0 + 16
= ±1
−4
4
Note that not all four choices (b, d) = (+1, +1), . . . , (−1, −1) satisfy ad − bc ≠ 0.
Subsection Five.II.3: Eigenvalues and Eigenvectors
Five.II.3.20
(a) This
0 =
∣ 10 − x −9
4 −2 − x∣ = (10 − x)(−2 − x) − (−36)
simplifies to the characteristic equation x 2 − 8x + 16 = 0. Because the equation factors into (x − 4) 2
there is only one eigenvalue λ 1 = 4.
(b) 0 = (1 − x)(3 − x) − 8 = x 2 − 4x − 5; λ 1 = 5, λ 2 = −1
(c) x 2 − 21 = 0; λ 1 = √ 21, λ 2 = − √ 21
(d) x 2 = 0; λ 1 = 0
(e) x 2 − 2x + 1 = 0; λ 1 = 1
Five.II.3.21 (a) The characteristic equation is (3 − x)(−1 − x) = 0. Its roots, the eigenvalues, are
λ 1 = 3 and λ 2 = −1. For the eigenvectors we consider this equation.
( ) ( ) ( 3 − x 0 b1 0
=
8 −1 − x b 2 0)
For the eigenvector associated with λ 1 = 3, we consider the resulting linear system.
0 · b 1 + 0 · b 2 = 0
8 · b 1 + −4 · b 2 = 0
The eigenspace is the set of vectors whose second component is twice the first component.
( ) ( ) ( ) ( )
b2 /2 ∣∣ 3 0 b2 /2 b2 /2
{ b
b 2 ∈ C}
= 3 ·
2 8 −1 b 2 b 2
(Here, the parameter is b 2 only because that is the variable that is free in the above system.) Hence,
this is an eigenvector associated with the eigenvalue 3.
(
1
2)
Finding an eigenvector associated with λ 2 = −1 is similar. This system
4 · b 1 + 0 · b 2 = 0
8 · b 1 + 0 · b 2 = 0
leads to the set of vectors whose first component is zero.
) ( ) ( )
0 ∣∣ 3 0 0
{(
b
b 2 ∈ C}
2 8 −1 b 2
And so this is an eigenvector associated with λ 2 .
(
0
1)
( )
0
= −1 ·
b 2
(b) The characteristic equation is
0 =
∣ 3 − x 2
−1 −x∣ = x2 − 3x + 2 = (x − 2)(x − 1)
and so the eigenvalues are λ 1 = 2 and λ 2 = 1. To find eigenvectors, consider this system.
(3 − x) · b 1 + 2 · b 2 = 0
−1 · b 1 − x · b 2 = 0
For λ 1 = 2 we get
1 · b 1 + 2 · b 2 = 0
−1 · b 1 − 2 · b 2 = 0