Page 97 - Linear Algebra Done Right

Chapter 5. Eigenvalues and Eigenvectors
84
Tv 1 ∈ span(v 1 ) ⊂ span(v 1 ,...,v k );
Tv 2 ∈ span(v 1 ,v 2 ) ⊂ span(v 1 ,...,v k );
. . .
Tv k ∈ span(v 1 ,...,v k ).
Thus if v is a linear combination of (v 1 ,...,v k ), then
Tv ∈ span(v 1 ,...,v k ).

In other words, span(v 1 ,...,v k ) is invariant under T, completing the
proof.

Now we can show that for each operator on a complex vector space,
there is a basis of the vector space with respect to which the matrix
of the operator has only 0’s below the diagonal. In Chapter 8 we will
improve even this result.

This theorem does not 5.13 Theorem: Suppose V is a complex vector space and T ∈L(V).
hold on real vector Then T has an upper-triangular matrix with respect to some basis of V.
spaces because the first
vector in a basis with Proof: We will use induction on the dimension of V. Clearly the
respect to which an desired result holds if dim V = 1.
operator has an Suppose now that dim V> 1 and the desired result holds for all
upper-triangular matrix complex vector spaces whose dimension is less than the dimension
must be an eigenvector of V. Let λ be any eigenvalue of T (5.10 guarantees that T has an
of the operator. Thus if eigenvalue). Let
an operator on a real U = range(T − λI).
vector space has no
Because T −λI is not surjective (see 3.21), dim U< dim V. Furthermore,
eigenvalues (we have
U is invariant under T. To prove this, suppose u ∈ U. Then
seen an example on
2
R ), then there is no Tu = (T − λI)u + λu.
basis with respect to
which the operator has Obviously (T − λI)u ∈ U (from the definition of U) and λu ∈ U. Thus
an upper-triangular the equation above shows that Tu ∈ U. Hence U is invariant under T,
matrix. as claimed.
Thus T| U is an operator on U. By our induction hypothesis, there
is a basis (u 1 ,...,u m ) of U with respect to which T| U has an upper-
triangular matrix. Thus for each j we have (using 5.12)

5.14 Tu j = (T| U )(u j ) ∈ span(u 1 ,...,u j ).