Page 359 - 35Linear Algebra

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Since this (spanning) set has three (linearly independent) elements,
3
the vector space of linear maps B → B has dimension 3. This is an
example of a general notion called the dual vector space.
   
a a
2
2
2
2 =
b
b
11. (a) d X d cos(ωt)   = −ω cos(ωt)   .
dt dt 2
c c
Hence
     
−a − b −1 −1 0 a
b
F = cos(ωt)  −a − 2b − c  = cos(ωt)  −1 −2 −1   
−b − c 0 −1 −1 c
 
a
2
= −ω cos(ωt)   ,
b
c
so
 
−1 −1 0
M =  −1 −2 −1  .
0 −1 −1
(b)
 
λ + 1 1 0

det  1 λ + 2 1  = (λ + 1) (λ + 2)(λ + 1) − 1 − (λ + 1)
0 1 λ + 1

= (λ + 1) (λ + 2)(λ + 1) − 2
2
= (λ + 1) λ + 3λ) = λ(λ + 1)(λ + 3)
so the eigenvalues are λ = 0, −1, −3.
For the eigenvectors, when λ = 0 we study:
     
−1 −1 0 1 1 0 1 0 −1
M − 0.I =  −1 −2 −1  ∼  0 −1 −1  ∼  0 1 1  ,
0 −1 −1 0 −1 −1 0 0 0
 
1
so  −1  is an eigenvector.
1
For λ = −1
   
0 −1 0 1 0 1
M − (−1).I =  −1 −1 −1  ∼  0 1 0  ,
0 −1 0 0 0 0

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