Page 105 - Linear Algebra Done Right

Chapter 5. Eigenvalues and Eigenvectors
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5.25
T u + α j Tu + β j u = 0.
We will complete the proof by showing that span(u, Tu), which clearly
has dimension 1 or 2, is invariant under T. To do this, consider a typical
element of span(u, Tu) of the form au+bTu, where a, b ∈ R. Then
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T(au + bTu) = aTu + bT u
= aTu − bα j Tu − bβ j u,
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where the last equality comes from solving for T u in 5.25. The equa-
tion above shows that T(au+bTu) ∈ span(u, Tu). Thus span(u, Tu)
is invariant under T, as desired.
We will need one new piece of notation for the next proof. Suppose
U and W are subspaces of V with

V = U ⊕ W.

Each vector v ∈ V can be written uniquely in the form
v = u + w,

P U,W is often called the where u ∈ U and w ∈ W. With this representation, define P U,W ∈L(V)
projection onto U with by
null space W. P U,W v = u.

You should verify that P U,W v = v if and only if v ∈ U. Interchanging
the roles of U and W in the representation above, we have P W,U v = w.
Thus v = P U,W v + P W,U v for every v ∈ V. You should verify that
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P U,W = P U,W ; furthermore range P U,W = U and null P U,W = W.
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We have seen an example of an operator on R with no eigenvalues.
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The following theorem shows that no such example exists on R .
5.26 Theorem: Every operator on an odd-dimensional real vector
space has an eigenvalue.

Proof: Suppose V is a real vector space with odd dimension. We
will prove that every operator on V has an eigenvalue by induction (in
steps of size 2) on the dimension of V. To get started, note that the
desired result obviously holds if dim V = 1.
Now suppose that dim V is an odd number greater than 1. Using
induction, we can assume that the desired result holds for all operators