Unformatted text preview: zero such that
x x x . Then 0. Now let z
x
x
x
. Then clearly, z is a
solution of x
P x , with z
. Furthermore, y
0 is also a solution,
with y
. By the uniqueness theorem, z
y
0 Hence
x x on the entire interval
9 . Let y 0 . Going in the other direction is trivial. be any solution of x
z x y P x x . It follows that
x is also a solution. Now let x y . Then the collection of vectors
x x x y constitutes
vectors, each with components. Based on the assertion in Prob.
Section
, these vectors are necessarily linearly dependent. That is, there are
constants
not all zero such that
x x x y x y , 0. From Prob. , we have
x x 0 for all
Now
, otherwise that would contradict the fact that the
first vectors are linearly independent. Hence
y x x x , and the assertion is true.
. Consider z
have x
z x
x x
x , and suppose that we also
x Based on the assumption, ________________________________________________________________________
page 364 —————————————————————————— CHAPTER 7. —— x x x 0. The collection of vectors
x
is linearly independe...
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 Spring '08
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 Linear Algebra, eigenvector

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