p(z) = c(z−λ1)· · ·(z−λm),
where c, λ1, . . . , λm ∈C and c6= 0.
Therefore,
0 =a0v+a1T v+a2T2v+· · ·+anTnv =p(T)v
=c(T −λ1I)(T −λ2I)· · ·(T −λmI)v,
and so at least one of the factors T −λjI must be non-injective. In other words, this λj is
an eigenvalue of T.
Note that the proof of Theorem 7.4.1 only uses basic concepts about linear maps, which is the same approach as in a popular textbook called Linear Algebra Done Right by Sheldon Axler. Many other textbooks rely on significantly more difficult proofs using concepts like the determinant and characteristic polynomial of a matrix. At the same time, it is often prefer- able to use the characteristic polynomial of a matrix in order to compute eigen-information of an operator; we discuss this approach in Chapter 8.
Note also that Theorem 7.4.1 does not hold for real vector spaces. E.g., as we saw in Example 7.2.2, the rotation operator R on R2 has no eigenvalues.
7.5
Upper triangular matrices
As before, let V be a complex vector space.
LetT ∈ L(V, V) and (v1, . . . , vn) be a basis for V. Recall that we can associate a matrix M(T) ∈ Cn×n to the operator T. By Theorem 7.4.1, we know that T has at least one
eigenvalue, say λ ∈ C. Let v1 6= 0 be an eigenvector corresponding to λ. By the Basis
92 CHAPTER 7. EIGENVALUES AND EIGENVECTORS
column ofM(T) with respect to this basis is λ 0 .. . 0 .
What we will show next is that we can find a basis ofV such that the matrixM(T) isupper triangular.
Definition 7.5.1. A matrix A = (aij) ∈ Fn×n is called upper triangular if aij = 0 for i > j.
Schematically, an upper triangular matrix has the form ∗ ∗ . .. 0 ∗ ,
where the entries ∗ can be anything and every entry below the main diagonal is zero. Here are two reasons why having an operatorT represented by an upper triangular matrix can be quite convenient:
1. the eigenvalues are on the diagonal (as we will see later);
2. it is easy to solve the corresponding system of linear equations by back substitution (as discussed in Section A.3).
The next proposition tells us what upper triangularity means in terms of linear operators and invariant subspaces.
Proposition 7.5.2. Suppose T ∈ L(V, V) and that (v1, . . . , vn) is a basis of V. Then the
following statements are equivalent:
1. the matrix M(T) with respect to the basis (v1, . . . , vn) is upper triangular;
7.5. UPPER TRIANGULAR MATRICES 93
3. span(v1, . . . , vk) is invariant under T for each k = 1,2, . . . , n.
Proof. The equivalence of Condition 1 and Condition 2 follows easily from the definition since Condition 2 implies that the matrix elements below the diagonal are zero.
Clearly, Condition 3 implies Condition 2. To show that Condition 2 implies Condition 3, note that any vector v ∈span(v1, . . . , vk) can be written asv =a1v1+· · ·+akvk. Applying T, we obtain
T v =a1T v1+· · ·+akT vk ∈span(v1, . . . , vk)
since, by Condition 2, each T vj ∈span(v1, . . . , vj) ⊂span(v1, . . . , vk) for j = 1,2, . . . , k and
since the span is a subspace of V.
The next theorem shows that complex vector spaces indeed have some basis for which the matrix of a given operator is upper triangular.
Theorem 7.5.3. Let V be a finite-dimensional vector space over C andT ∈ L(V, V). Then there exists a basis B for V such that M(T) is upper triangular with respect to B.
Proof. We proceed by induction on dim(V). If dim(V) = 1, then there is nothing to prove. Hence, assume that dim(V) =n >1 and that we have proven the result of the theorem for all T ∈ L(W, W), where W is a complex vector space with dim(W) ≤ n − 1. By Theorem 7.4.1, T has at least one eigenvalueλ. Define
U = range (T −λI),
and note that
1. dim(U)<dim(V) = n sinceλis an eigenvalue ofT and hence T−λI is not surjective; 2. U is an invariant subspace of T since, for all u∈U, we have
T u= (T −λI)u+λu,
which implies that T u∈U since (T −λI)u∈range (T −λI) =U and λu∈U.
Therefore, we may consider the operator S = T|U, which is the operator obtained by re-
94 CHAPTER 7. EIGENVALUES AND EIGENVECTORS
of U with m ≤n−1 such that M(S) is upper triangular with respect to (u1, . . . , um). This
means that
T uj =Suj ∈span(u1, . . . , uj), for all j = 1,2, . . . , m.
Extend this to a basis (u1, . . . , um, v1, . . . , vk) of V. Then
T vj = (T −λI)vj +λvj, for all j = 1,2, . . . , k.
Since (T −λI)vj ∈range (T −λI) = U = span(u1, . . . , um), we have that T vj ∈span(u1, . . . , um, v1, . . . , vj), for all j = 1,2, . . . , k.
Hence, T is upper triangular with respect to the basis (u1, . . . , um, v1, . . . , vk).
The following are two very important facts about upper triangular matrices and their associated operators.
Proposition 7.5.4. Suppose T ∈ L(V, V) is a linear operator and that M(T) is upper triangular with respect to some basis of V. Then
1. T is invertible if and only if all entries on the diagonal of M(T) are non-zero. 2. The eigenvalues of T are precisely the diagonal elements of M(T).
Proof of Proposition 7.5.4, Part 1. Let (v1, . . . , vn) be a basis of V such that
M(T) = λ1 ∗ . .. 0 λn
is upper triangular. The claim is that T is invertible if and only if λk 6= 0 for all k =
1,2, . . . , n. Equivalently, this can be reformulated as follows: T is not invertible if and only if λk = 0 for at least one k ∈ {1,2, . . . , n}.
Suppose λk= 0. We will show that this implies the non-invertibility of T. If k= 1, this
is obvious since thenT v1 = 0, which implies thatv1 ∈null (T) so that T is not injective and
hence not invertible. So assume thatk > 1. Then
7.5. UPPER TRIANGULAR MATRICES 95 since T is upper triangular and λk = 0. Hence, we may define S = T|span(v1,...,vk) to be the restriction of T to the subspace span(v1, . . . , vk) so that
S : span(v1, . . . , vk)→span(v1, . . . , vk−1).
The linear map S is not injective since the dimension of the domain is larger than the dimension of its codomain, i.e.,
dim(span(v1, . . . , vk)) = k > k−1 = dim(span(v1, . . . , vk−1)).
Hence, there exists a vector 0 6=v ∈ span(v1, . . . , vk) such that Sv = T v = 0. This implies
that T is also not injective and therefore not invertible.
Now suppose that T is not invertible. We need to show that at least one λk = 0. The
linear mapT not being invertible implies thatT is not injective. Hence, there exists a vector 06=v ∈V such that T v = 0, and we can write
v =a1v1+· · ·+akvk
for some k, whereak 6= 0. Then
0 =T v = (a1T v1+· · ·+ak−1T vk−1) +akT vk. (7.2)
SinceT is upper triangular with respect to the basis (v1, . . . , vn), we know thata1T v1+· · ·+
ak−1T vk−1 ∈span(v1, . . . , vk−1). Hence, Equation (7.2) shows thatT vk ∈span(v1, . . . , vk−1),
which implies that λk = 0.
Proof of Proposition 7.5.4, Part 2. Recall that λ ∈ F is an eigenvalue of T if and only if the operator T −λI is not invertible. Let (v1, . . . , vn) be a basis such that M(T) is upper
triangular. Then M(T −λI) = λ1−λ ∗ . .. 0 λn−λ .
96 CHAPTER 7. EIGENVALUES AND EIGENVECTORS