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MAT067 University of California, Davis Winter 2007
Eigenvalues and Eigenvectors
Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 12, 2007)
In this section we are going to study linear maps T : V → V from a vector space to itself. These linear maps are also called operators on V. We are interested in the question when there is a basis for V such that T has a particularly nice form, like being diagonal or upper triangular. This quest leads us to the notion of eigenvalues and eigenvectors of linear operators, which is one of the most important concepts in linear algebra and beyond. For example, quantum mechanics is based on the study of eigenvalues and eigenvectors of operators.
1 Invariant subspaces
To begin our study we will look at subspaces U of V that have special properties under an operator T ∈ L(V, V ).
Definition 1. Let V be a finite-dimensional vector space over F with dim V ≥ 1, and let T ∈ L(V, V ) be an operator in V. Then a subspace U ⊂ V is called an invariant subspace under T if T u ∈ U for all u ∈ U.
That is if T U = {T u | u ∈ U} ⊂ U.
Example 1. The subspaces null T and range T are invariant subspaces under T. To see this, let u ∈ null T. This means that T u = 0. But since 0 ∈ null T this implies that T u = 0 ∈ null T. Similarly, let u ∈ range T. Since T v ∈ range T for all v ∈ V , we certainly also have that T u ∈ range T.
An important special case is the case of one-dimensional invariant subspaces of an oper- ator T ∈ L(V, V ). If dim U = 1, then there exists a nonzero vector u ∈ V such that
U = {au | a ∈ F}.
Copyright ©c 2007 by the authors. These lecture notes may be reproduced in their entirety for non- commercial purposes.
2 EIGENVALUES 2
In this case we must have T u = λu for some λ ∈ F.
This motivates the definition of eigenvectors and eigenvalues of a linear operator T.
2 Eigenvalues
Definition 2. Let T ∈ L(V, V ). Then λ ∈ F is an eigenvalue of T if there exists a nonzero vector u ∈ V such that T u = λu.
The vector u is called the eigenvector (with eigenvalue λ) of T.
Finding the eigenvalues and eigenvectors of linear operators is one of the most important problems in linear algebra. We will see later that they have many uses and applications. For example all of quantum mechanics is based on eigenvalues and eigenvectors of operators.
Example 2.
- Let T be the zero map defined by T (v) = 0 for all v ∈ V. Then every vector u = 0 is an eigenvector of T with eigenvalue 0.
- Let I be the identity map defined by I(v) = v for all v ∈ V. Then every vector u = 0 is an eigenvector of T with eigenvalue 1.
- The projection P : R^3 → R^3 defined by P (x, y, z) = (x, y, 0) has eigenvalues 0 and
- The vector (0, 0 , 1) is an eigenvector with eigenvalue 0 and (1, 0 , 0) and (0, 1 , 0) are eigenvectors with eigenvalue 1.
- Take the operator R : F^2 → F^2 defined by R(x, y) = (−y, x). When F = R, then R can be interpreted as counterclockwise rotation by 90^0. From this interpretation it is clear that no vector in R^2 is left invariant (up to a scalar multiplication). Hence for F = R the operator R has no eigenvalues. For F = C the situation is different! Then λ ∈ C is an eigenvalue if R(x, y) = (−y, x) = λ(x, y), so that y = −λx and x = λy. This implies that y = −λ^2 y or λ^2 = −1. The solutions are hence λ = ±i. One can check that (1, −i) is an eigenvector with eigenvalue i and (1, i) is an eigenvector with eigenvalue −i.
Eigenspaces are important examples of invariant subspaces. Let T ∈ L(V, V ) and let λ ∈ F be an eigenvalue of T. Then
Vλ = {v ∈ V | T v = λv}
3 DIAGONAL MATRICES 4
3 Diagonal matrices
Note that if T has n = dim V distinct eigenvalues, then there exists a basis (v 1 ,... , vn) of V such that T vj = λj vj for all j = 1, 2 ,... , n.
Then any v ∈ V can be written as a linear combination of v 1 ,... , vn as v = a 1 v 1 + · · · + anvn. Applying T to this we obtain
T v = λ 1 a 1 v 1 + · · · + λnanvn.
Hence the vector
M(v) =
a 1 .. . an
is mapped to
M(T v) =
λ 1 a 1 .. . λnan
This means that the matrix M(T ) for T with respect to the basis of eigenvectors (v 1 ,... , vn) is diagonal
M(T ) =
λ 1 0
... 0 λn
We summarize the results of the above discussion in the following Proposition.
Proposition 3. If T ∈ L(V, V ) has dim V distinct eigenvalues, then M(T ) is diagonal with respect to some basis of V. Moreover, V has a basis consisting of eigenvectors of T.
4 Existence of eigenvalues
In what follows we want to study the question of when an operator T has any eigenvalue. To answer this question we will use polynomials p(z) ∈ P(F) evaluated on operators T ∈ L(V, V ) or equivalently on square matrices Fn×n. More explicitly, for a polynomial
p(z) = a 0 + a 1 z + · · · + akzk
4 EXISTENCE OF EIGENVALUES 5
we can associate the operator
p(T ) = a 0 IV + a 1 T + · · · + akT k.
Note that for p, q ∈ P(F) we have
(pq)(T ) = p(T )q(T ) = q(T )p(T ).
The results of this section will be for complex vector spaces. The reason for this is that the proof of the existence of eigenvalues relies on the Fundamental Theorem of Algebra, which makes a statement about the existence of zeroes of polynomials over the complex numbers.
Theorem 4. Let V = { 0 } be a finite-dimensional vector space over C and T ∈ L(V, V ). Then T has at least one eigenvalue.
Proof. Let v ∈ V , v = 0 and consider
(v, T v, T 2 v,... , T nv),
where n = dim V. Since the list contains n + 1 vectors, it must be linearly dependent. Hence there exist scalars a 0 , a 1 ,... , an ∈ C, not all zero, such that
0 = a 0 v + a 1 T v + a 2 T 2 v + · · · + anT nv.
Let m be largest such that am = 0. Since v = 0 we must have m > 0 (but possibly m = n). Consider the polynomial
p(z) = a 0 + a 1 z + · · · + amzm^ = c(z − λ 1 ) · · · (z − λm),
where c, λ 1 ,... , λm ∈ C and c = 0. Note that the polynomial factors because of the Funda- mental Theorem of Algebra, which says that every polynomial over C has at least one zero. Call this zero λ 1. Then p(z) = (z − λ 1 )˜p(z) where ˜p(z) is a polynomial of degree m − 1. Continue the process until p(z) is completely factored. Therefore
0 = a 0 v + a 1 T v + a 2 T 2 v + · · · + anT nv = p(T )v = c(T − λ 1 I)(T − λ 2 I) · · · (T − λmI)v,
so that at least one of the factors T − λj I must be noninjective. In other words, this λj is an eigenvalue of T.
Note that the proof of this Theorem only uses basic concepts about linear maps, not
5 UPPER TRIANGULAR MATRICES 7
- the matrix M(T ) with respect to the basis (v 1 ,... , vn) is upper triangular;
- T vk ∈ span(v 1 ,... , vk) for each k = 1, 2 ,... , n;
- span(v 1 ,... , vk) is invariant under T for each k = 1, 2 ,... , n.
Proof. The equivalence of 1 and 2 follows easily from the definition since 2 implies that the matrix elements below the diagonal are zero. Obviously 3 implies 2. To show that 2 implies 3 note that any vector v ∈ span(v 1 ,... , vk) can be written as v = a 1 v 1 + · · · + akvk. Applying T we obtain
T v = a 1 T v 1 + · · · + akT vk ∈ span(v 1 ,... , vk)
since by 2 each T vj ∈ span(v 1 ,... , vj ) ⊂ span(v 1 ,... , vk) for j = 1, 2 ,... , k and the span is a subspace of V and hence closed under addition and scalar multiplication.
The next theorem shows that complex vector spaces indeed have some basis for which the matrix of a given operator is upper triangular.
Theorem 6. Let V be a finite-dimensional vector space over C and T ∈ L(V, V ). Then there exists a basis for V such that M(T ) is upper triangular with respect to this basis.
Proof. We proceed by induction on dim V. If dim V = 1 there is nothing to prove. Hence assume that dim V = n > 1 and we have proved 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 4 T has at least one eigenvalue λ. Define
U = range (T − λI).
Note that
- dim U < dim V = n since λ is an eigenvalue of T and hence T − λI is not surjective;
- 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 , being the operator T restricted to the subspace U. By induction hypothesis there exists a basis (u 1 ,... , um) of U with m ≤ n − 1 such that M(S) is upper triangular with respect to (u 1 ,... , um). This means that
T uj = Suj ∈ span(u 1 ,... , uj ) for all j = 1, 2 ,... , m.
5 UPPER TRIANGULAR MATRICES 8
Extend this to a basis (u 1 ,... , um, v 1 ,... , 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(u 1 ,... , um), we have that
T vj ∈ span(u 1 ,... , um, v 1 ,... , vj ) for all j = 1, 2 ,... , k.
Hence T is upper triangular with respect to the basis (u 1 ,... , um, v 1 ,... , vk).
There are two very important facts about upper triangular matrices and their associated operators.
Proposition 7. Suppose T ∈ L(V, V ) is a linear operator and M(T ) is upper triangular with respect to some basis of V. Then
- T is invertible if and only if all entries on the diagonal of M(T ) are nonzero.
- The eigenvalues of T are precisely the diagonal elements of M(T ).
Proof of Proposition 7 Part 1. Let (v 1 ,... , vn) be a basis of V such that
M(T ) =
λ 1 ∗
... 0 λn
is upper triangular. The claim is that T is invertible if and only if λk = 0 for all k = 1 , 2 ,... , n. Equivalently, this can be reformulated as 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 then T is not invertible. If k = 1, this is obvious since then T v 1 = 0, which implies that v 1 ∈ null T so that T is not injective and hence not invertible. So assume that k > 1. Then
T vj ∈ span(v 1 ,... , vk− 1 ) for all j ≤ k
since T is upper triangular and λk = 0. Hence we may define S = T |span(v 1 ,...,vk) to be the restriction of T to the subspace span(v 1 ,... , vk)
S : span(v 1 ,... , vk) → span(v 1 ,... , vk− 1 ).
The linear map S is not injective since the dimension of the domain is bigger than the dimension of its codomain
dim span(v 1 ,... , vk) = k > k − 1 = dim span(v 1 ,... , vk− 1 ).
6 DIAGONALIZATION OF 2 × 2 MATRICES AND APPLICATIONS 10
eigenvalue λ ∈ F if and only if the system of linear equations
(a − λ)v 1 + bv 2 = 0 cv 1 + (d − λ)v 2 = 0
has a non-trivial solution. Moreover, the system of equations (3) has a non-trivial solution if and only if the polynomial p(λ) = (a − λ)(d − λ) − bc evaluates to zero (see Homework 7, Problem 1). In other words, the eigenvalues for T are exactly the λ ∈ F for which p(λ) = 0, and the
eigenvectors for T associated to an eigenvalue λ are exactly the non-zero vectors v =
[
v 1 v 2
]
F^2 that satisfy the system of equations (3).
Example 3. Let A =
[
]
. Then p(λ) = (− 2 − λ)(2 − λ) − (−1)(5) = λ^2 + 1, which
is equal to zero exactly when λ = ±i. Moreover, if λ = i, then the system of equations (3) becomes (− 2 − i)v 1 − v 2 = 0 5 v 1 + (2 − i)v 2 = 0
which is satisfied by any vector v =
[
v 1 v 2
]
∈ C^2 such that v 2 = (− 2 − i)v 1. Similarly, if
λ = −i, then the system of equations (3) becomes
(−2 + i)v 1 − v 2 = 0 5 v 1 + (2 + i)v 2 = 0
which is satisfied by any vector v =
[
v 1 v 2
]
∈ C^2 such that v 2 = (−2 + i)v 1.
It follows that, given A =
[
]
, the linear operator on C^2 defined by T (v) = Av
has eigenvalues λ = ±i, with associated eigenvectors as described above.
Example 4. Take the rotation Rθ : R^2 → R^2 by an angle θ ∈ [0, 2 π) given by the matrix
Rθ =
[
cos θ − sin θ sin θ cos θ
]
Then we obtain the eigenvalues by solving the polynomial equation
p(λ) = (cos θ − λ)^2 + sin^2 θ = λ^2 − 2 λ cos θ + 1 = 0,
6 DIAGONALIZATION OF 2 × 2 MATRICES AND APPLICATIONS 11
where we used that sin^2 θ + cos^2 θ = 1. Solving for λ in C we obtain
λ = cos θ ±
cos^2 θ − 1 = cos θ ±
− sin^2 θ = cos θ ± i sin θ = e±iθ.
We see that as an operator over the real vector space R^2 , the operator Rθ only has eigenvalues
when θ = 0 or θ = π. However, if we interpret the vector
[
x 1 x 2
]
∈ R^2 as a complex number
z = x 1 + ix 2 , then z is an eigenvector if Rθ : C → C maps z → λz = e±iθz. We already saw in the lectures on complex numbers that multiplication by e±iθ^ corresponds to rotation by the angle ±θ.
