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BEGINNINGS OF EIGENSTUFF
INNA ZAKHAREVICH
In these notes, let V be a vector space over a field F , and T be a linear transformation V → V .
Definition 0.1. An eigenvalue λ ∈ F of T is an element of F such that there exists a nonzero
v ∈ V such that
T v = λv.
In this case, v is called an eigenvector with eigenvalue λ.
In Hoffman& Kunze, eigenvalues are called “characteristic values.”
Note that λ is an eigenvalue of T if and only if the linear transformation
T − λI
is not injective.
Examples. Let F = R.
(1) If V = R2 and
T =
2 0
0 3
0
T =
1 1
0 1
00
T =
0 1
−1 0
then T has two eigenvalues 2 and 3 with associated eigenvectors e1 and e2 , T 0 has one
eigenvalue 1 with eigenvector e1 and T 00 has no eigenvalues.
(2) If V = C ∞ (R, R) and T = d/dx then 0, 1 and 2 are eigenvalues of T , with associated
eigenvectors C, ex and e2x (for C a constant).
(3) If T is a linear transformation V → V and v is an eigenvector of T with eigenvalue λ, then
cv is an eigenvector of T with eigenvalue λ for all c ∈ F × .
Theorem 0.2. If V is finite dimensional and F is algebraically closed then T has an eigenvalue.
Proof. Let dim V = n. Let v be any nonzero vector and consider the set
{v, T v, . . . , T n v}.
Then this set has size n + 1 and so it is linearly dependent. Thus there exist a0 , . . . , an , not all
zero, such that
a0 v + a1 T v + · · · + an T n v = 0.
Let
p(x) = a0 + a1 x + · · · + an xn .
Then p(T ) is a linear transformation V → V and
p(T )v = 0.
Since F is algebraically closed, we can write
p(x) = (x − λ1 ) · · · (x − λm ),
where m = deg p. Then
p(T ) = (T − λ1 I) · · · (T − λm I),
and so
(T − λ1 I) · · · (T − λm I)v = 0.
1
2
INNA ZAKHAREVICH
Since v is nonzero this means that the composition of the linear transformations T −λ1 I, . . . , T −λm I
is not injective, so at least one of these must not be injective. If T − λj I is not injective and
w ∈ ker(T − λj I) then
(T − λj I)w = 0 =⇒ T w = λj w.
Thus w is an eigenvector of T with eigenvalue λj , and we are done.
Proposition 0.3. Suppose that {vi }i∈I is a set of eigenvectors of T with associated eigenvalues
{λi }i∈I . If λi 6= λj whenever i 6= j then the set {vi }i∈I is linearly independent.
Proof. We will prove that for all n less than or equal to the cardinality of I (which may be all n,
if I is infinite) that any subset {vi1 , . . . , vin } of size n of {vi }i∈I is linearly independent. This is
clearly true for n = 1, since eigenvectors are nonzero by definition. Now suppose that it’s true for
any subset of size at most n − 1, and consider a subset of size n.
Suppose that the subset is linearly dependent; then there exist a1 , . . . , an ∈ F such that
a1 vi1 + · · · + an vin = 0.
Note that all ai must be nonzero; if for example a1 were zero then the set {vi2 , . . . , vin } would be
linearly dependent, contradicting the inductive hypothesis. Apply T ot each side of the equation;
this gives us the equation
λi1 a1 vi1 + · · · + λin an vin = 0.
All of the λ’s must also be nonzero. Indeed, at most one is zero, and if, for example, λi1 = 0
then we get that {vi2 , . . . , vin } is linearly dependent, again contradicting the inductive hypothesis.
However, if λi1 6= 0 then we can divide both sides of the equation by it to get
λi
λi
a1 vi1 + 2 a2 vi2 + · · · + n an vin = 0.
λi1
λ i1
Subtracting this equation from the original one, we get the equation
λi
λi
(1 − 2 )a2 v2 + · · · + (1 − n )an vn = 0.
λ i1
λi1
However, since the λ’s are distinct, all coefficients are nonzero and we see that {vi2 , . . . , vin } is
linearly dependent. This contradicts our induction hypothesis, and we are done.