Page 106 - Linear Algebra Done Right

Invariant Subspaces on Real Vector Spaces
on all real vector spaces with dimension 2 less than dim V. Suppose
T ∈L(V). We need to prove that T has an eigenvalue. If it does, we are
done. If not, then by 5.24 there is a two-dimensional subspace U of V 93
that is invariant under T. Let W be any subspace of V such that
V = U ⊕ W;
2.13 guarantees that such a W exists.
Because W has dimension 2 less than dim V, we would like to apply
our induction hypothesis to T| W . However, W might not be invariant
under T, meaning that T| W might not be an operator on W. We will
compose with the projection P W,U to get an operator on W. Specifically,
define S ∈L(W) by
Sw = P W,U (Tw)
for w ∈ W. By our induction hypothesis, S has an eigenvalue λ.We
will show that this λ is also an eigenvalue for T.
Let w ∈ W be a nonzero eigenvector for S corresponding to the
eigenvalue λ; thus (S − λI)w = 0. We would be done if w were an
eigenvector for T with eigenvalue λ; unfortunately that need not be
true. So we will look for an eigenvector of T in U + span(w).To do
that, consider a typical vector u + aw in U + span(w), where u ∈ U
and a ∈ R. We have

(T − λI)(u + aw) = Tu − λu + a(Tw − λw)
= Tu − λu + a(P U,W (Tw) + P W,U (Tw) − λw)
= Tu − λu + a(P U,W (Tw) + Sw − λw)
= Tu − λu + aP U,W (Tw).

Note that on the right side of the last equation, Tu ∈ U (because U
is invariant under T), λu ∈ U (because u ∈ U), and aP U,W (Tw) ∈ U
(from the definition of P U,W ). Thus T − λI maps U + span(w) into U.
Because U + span(w) has a larger dimension than U, this means that
(T − λI)| U+span(w) is not injective (see 3.5). In other words, there exists
a nonzero vector v ∈ U + span(w) ⊂ V such that (T − λI)v = 0. Thus
T has an eigenvalue, as desired.