Page 104 - Linear Algebra Done Right

Invariant Subspaces on Real Vector Spaces
Invariant Subspaces on Real Vector Spaces
We know that every operator on a complex vector space has an eigen- 91
value (see 5.10 for the precise statement). We have also seen an example
showing that the analogous statement is false on real vector spaces. In
other words, an operator on a nonzero real vector space may have no
invariant subspaces of dimension 1. However, we now show that an
invariant subspace of dimension 1 or 2 always exists.

5.24 Theorem: Every operator on a finite-dimensional, nonzero, real
vector space has an invariant subspace of dimension 1 or 2.

Proof: Suppose V is a real vector space with dimension n> 0 and
T ∈L(V). Choose v ∈ V with v = 0. Then

2
n
(v, Tv, T v,...,T v)
cannot be linearly independent because V has dimension n and we have
n + 1 vectors. Thus there exist real numbers a 0 ,...,a n , not all 0, such
that
n
0 = a 0 v + a 1 Tv +· · ·+ a n T v.
Make the a’s the coefficients of a polynomial, which can be written in
factored form (see 4.14) as

a 0 + a 1 x + ··· + a n x n
2
2
= c(x − λ 1 )...(x − λ m )(x + α 1 x + β 1 )...(x + α M x + β M ), Here either m or M
might equal 0.
where c is a nonzero real number, each λ j , α j , and β j is real, m+M ≥ 1,
and the equation holds for all x ∈ R. We then have
n
0 = a 0 v + a 1 Tv +· · ·+ a n T v
n
= (a 0 I + a 1 T + ··· + a n T )v
2
2
= c(T − λ 1 I)...(T − λ m I)(T + α 1 T + β 1 I)...(T + α M T + β M I)v,
which means that T − λ j I is not injective for at least one j or that
2
(T + α j T + β j I) is not injective for at least one j.If T − λ j I is not
injective for at least one j, then T has an eigenvalue and hence a one-
dimensional invariant subspace. Let’s consider the other possibility. In
2
other words, suppose that (T + α j T + β j I) is not injective for some j.
Thus there exists a nonzero vector u ∈ V such that