Page 90 - Linear Algebra Done Right

Invariant Subspaces
of invariant subspaces other than {0} and V because null T may equal
{0} and range T may equal V (this happens when T is invertible).
We will return later to a deeper study of invariant subspaces. Now 77
we turn to an investigation of the simplest possible nontrivial invariant
subspaces—invariant subspaces with dimension 1.
How does an operator behave on an invariant subspace of dimen-
sion 1? Subspaces of V of dimension 1 are easy to describe. Take any
nonzero vector u ∈ V and let U equal the set of all scalar multiples
of u:
5.2 U ={au : a ∈ F}.

Then U is a one-dimensional subspace of V, and every one-dimensional These subspaces are
subspace of V is of this form. If u ∈ V and the subspace U defined loosely connected to
by 5.2 is invariant under T ∈L(V), then Tu must be in U, and hence the subject of Herbert
there must be a scalar λ ∈ F such that Tu = λu. Conversely, if u Marcuse’s well-known
is a nonzero vector in V such that Tu = λu for some λ ∈ F, then the book One-Dimensional
subspace U defined by 5.2 is a one-dimensional subspace of V invariant Man.
under T.
The equation

5.3 Tu = λu,

which we have just seen is intimately connected with one-dimensional
invariant subspaces, is important enough that the vectors u and scalars
λ satisfying it are given special names. Specifically, a scalar λ ∈ F
is called an eigenvalue of T ∈L(V) if there exists a nonzero vector The regrettable word
u ∈ V such that Tu = λu. We must require u to be nonzero because eigenvalue is
with u = 0 every scalar λ ∈ F satisfies 5.3. The comments above show half-German,
that T has a one-dimensional invariant subspace if and only if T has half-English. The
an eigenvalue. German adjective eigen
The equation Tu = λu is equivalent to (T − λI)u = 0, so λ is an means own in the sense
eigenvalue of T if and only if T − λI is not injective. By 3.21, λ is an of characterizing some
eigenvalue of T if and only if T − λI is not invertible, and this happens intrinsic property.
if and only if T − λI is not surjective. Some mathematicians
Suppose T ∈L(V) and λ ∈ F is an eigenvalue of T. A vector u ∈ V use the term
is called an eigenvector of T (corresponding to λ)if Tu = λu. Because characteristic value
5.3 is equivalent to (T − λI)u = 0, we see that the set of eigenvectors instead of eigenvalue.
of T corresponding to λ equals null(T − λI). In particular, the set of
eigenvectors of T corresponding to λ is a subspace of V.