Math 121B — Linear Algebra
Neil Donaldson
Fall 2022
Linear Algebra, Stephen Friedberg, Arnold Insel & Lawrence Spence, 4th Ed 2003, Prentice Hall.
Review from 121A
We begin by recalling a few basic notions and notations.
Vector Spaces Bold-face v denotes a vector in a vector space V over a field F. A vector space is closed
under vector addition and scalar multiplication
∀v
1
, v
2
∈ V, ∀λ
1
, λ
2
∈ F, λ
1
v
1
+ λ
2
v
2
∈ V
Examples. Here are four (families of) vector spaces over the field R.
• R
2
= {xi + yj : x, y ∈ R} = {
(
x
y
)
: x, y ∈ R} is a vector space over the field R.
• P
n
(R ); polynomials with degree ≤ n and coefficients in R
• P(R); polynomials over R with any degree.
• C(R); continuous functions from R to R.
Linear Combinations and Spans Let β ⊆ V be a subset of a vector space V over F. A linear
combination of vectors in β is any finite sum
λ
1
v
1
+ · · · + λ
n
v
n
where λ
j
∈ F and v
j
∈ β. The span of β comprises all linear combinations: this is a subspace of V.
Bases and Co-ordinates A set β ⊆ V is a basis of V if it has two properties:
Linear Independence Any linear combination yielding the zero vector is trivial; for distinct v
j
∈ β,
λ
1
v
1
+ · · · + λ
n
v
n
= 0 =⇒ ∀j, λ
j
= 0
Spanning Set V = Span β; every vector in V is a (finite!) linear combination of elements of β.
Theorem. β is a basis of V ⇐⇒ every v ∈ V is a unique linear combination of elements of β.
The cardinality of all basis sets is identical; this is the dimension dim
F
V.
1
Example. P
2
(R ) has standard basis β = {1, x, x
2
}: every degree ≤ 2 polynomial is a unique as
a linear combination p(x) = a + bx + cx
2
and so dim P
2
(R ) = 3. The real numbers a, b, c are the
co-ordinates of p with respect to β; the co-ordinate vector of p is written
[p]
β
=
a
b
c
Linearity and Linear Maps A function T : V → W between vector spaces V, W over the same field
F is (F-)linear if it respects the linearity properties of V, W
∀v
1
, v
2
∈ V, ∀λ
1
, λ
2
∈ F, T( λ
1
v
1
+ λ
2
v
2
) = λ
1
T(v
1
) + λ
2
T(v
2
)
We write L(V, W) for the set (indeed vector space!) of linear maps from V to W: this is shortened to
L(V) if V = W. An isomorphism is an invertible/bijective linear map.
Theorem. If dim
F
V = n and β is a basis of V, then the co-ordinate map v 7→ [v]
β
is an isomorphism
of vector spaces V → F
n
.
Matrices and Linear Maps If V, W are finite-dimensional, then any linear map T : V → W can be
described using matrix multiplication.
Example. If A =
2 −1
0 −1
−4 3
, then the linear map L
A
: R
2
→ R
3
(left-multiplication by A) is
L
A
x
y
=
2 −1
0 −1
−4 3
x
y
=
2x − y
−y
3y − 4x
The linear map in fact defines the matrix A; we recover the columns of the matrix by feeding the
standard basis vectors to the linear map.
2
0
−4
= L
A
1
0
−1
−1
3
= L
A
0
1
More generally, if T ∈ L(V, W) and β = {v
1
, . . . , v
n
} and γ = {w
1
, . . . , w
m
} are bases of V, W
respectively, then the matrix of T with respect to β and γ is
[T]
γ
β
=
[T(v
1
)]
γ
· · · [T(v
n
)]
γ
∈ M
m×n
(F )
whose j
th
column is obtained by feeding the j
th
basis vector of β to T and taking its co-ordinate vector
with respect to γ. This fits naturally with the co-ordinate isomorphisms
T(v) = w ⇐⇒ [ T]
γ
β
[v]
β
= [w]
γ
2
There are two special cases when V = W:
• If β = γ, then we simply write [T]
β
instead of [T]
β
β
.
• If T = I is the identity map, then Q
γ
β
:= [I]
γ
β
is the change of co-ordinate matrix from β to γ.
Being able to convert linear maps into matrix multiplication is a central skill in linear algebra. Test
your comfort by working through the following; if everything feels familiar, you should consider
yourself in a good place as far as pre-requisites are concerned!
Example. Let T : P
2
(R ) → P
1
(R ) be the linear map defined by differentiation
T(a + bx + cx
2
) = b + 2cx (∗)
The standard bases of P
2
(R ) and P
1
(R ) are, respectively, β = {1, x, x
2
} and γ = {1, x}. Observe that
[T(1)]
γ
= [0]
γ
=
0
0
, [T(x)]
γ
= [1]
γ
=
1
0
, [T(x
2
)]
γ
= [2x]
γ
=
0
2
=⇒ [T]
γ
β
=
[T(1)]
γ
[T(x)]
γ
[T(x
2
)]
γ
=
0 1 0
0 0 2
Written in co-ordinates, we see the original linear map (∗)
T(a + bx + cx
2
)
γ
= [T]
γ
β
[a + bx + cx
2
]
β
=
0 1 0
0 0 2
a
b
c
=
b
2c
=
[
b + 2cx
]
γ
(†)
1. η = {1 + x, x + x
2
, x
2
+ 1} is also a basis of P
2
(R ). Show that
[T]
γ
η
=
1 1 0
0 2 2
2. As in (†) above, the matrix multiplication
1 1 0
0 2 2
a
b
c
=
a + b
2b + 2c
corresponds to an equation [T(p)]
γ
= [T]
γ
η
[p]
η
for some polynomial p(x); what is p(x) in terms
of a, b, c?
3. Find the change of co-ordinate matrix Q
β
η
and check that the matrices of T are related by
[T]
γ
η
= [T]
γ
β
Q
β
η
3
1 Diagonalizability & the Cayley–Hamilton Theorem
1.1 Eigenvalues, Eigenvectors & Diagonalization (Review)
Definition 1.1. Suppose V is a vector space over F and T ∈ L(V). A non-zero v ∈ V is an eigenvector
of T with eigenvalue λ ∈ F (together an eigenpair) if
T(v) = λv
For matrices, the eigenvalues/vectors of A ∈ M
n
(F ) are precisely those of L
A
∈ L(F
n
).
Suppose λ is an eigenvalue of T;
1. The eigenspace of λ is the nullspace E
λ
:= N (T − λI).
2. The geometric multiplicity of λ of the dimension dim E
λ
.
We say that T is diagonalizable if there exists a basis of eigenvectors; an eigenbasis.
We start by recalling a couple of basic facts, the first of which is easily proved by induction.
Lemma 1.2. If v
1
, . . . , v
k
are eigenvectors corresponding to distinct eigenvalues, then {v
1
, . . . , v
k
}
is linearly independent.
Moreover, if dim
F
V = n and T ∈ L(V) has n distinct eigenvalues, then T is diagonalizable.
Eigenvalues and Eigenvectors in finite dimensions
If dim
F
V = n and ϵ is a basis, then the eigenvector definition is equivalent to a matrix equation
[T]
ϵ
[v]
ϵ
= λ[v]
ϵ
In such a situation, T being diagonalizable means ∃β such that [T]
β
is a diagonal matrix
[T]
β
=
λ
1
0 · · · 0
0 λ
2
.
.
.
.
.
.
.
.
.
0
0 · · · 0 λ
n
Thankfully there is systematic way to find eigenvalues and eigenvectors in finite-dimensions:
1. Choose any basis ϵ of V and compute the matrix A = [T]
ϵ
∈ M
n
(F ).
2. Observe that
λ ∈ F is an eigenvalue ⇐⇒ ∃[v]
ϵ
∈ F
n
\ {0} such that A[v]
ϵ
= λ[v]
ϵ
⇐⇒ ∃[v]
ϵ
∈ F
n
\ {0} such that
(
A − λI
)
[v]
ϵ
= 0
⇐⇒ det
(
A − λI
)
= 0
This last is a degree n polynomial equation whose roots are the eigenvalues.
3. For each eigenvalue λ
j
, compute the eigenspace E
λ
j
= N (T − λ
j
I) to find the eigenvectors.
Remember that E
λ
j
is a subspace of the original vector space V, so translate back if necessary!
4
Definition 1.3. The characteristic polynomial of T ∈ L(V) is the degree-n polynomial
p(t) := det(T − tI)
The eigenvalues of T are precisely the solutions to the characteristic equation p(t) = 0.
Examples 1.4. 1. A =
0 −1
1 0
has characteristic polynomial p(t) = t
2
+ 1 = (t + i)(t − i). As a
linear map L
A
∈ L(R
2
), A has no eigenvalues and no eigenvectors!
As a linear map L
A
∈ L(C
2
), we have two eigenvalues ±i. Indeed
(A − iI)v =
−i −1
1 −i
v =⇒ E
i
= Span
i
1
and similarly E
−i
= Span
−i
1
. We therefore have an eigenbasis β = {
i
1
,
−i
1
} (of C
2
), with
respect to which
[L
A
]
β
=
i 0
0 −i
2. Let T ∈ L(P
2
(R )) be defined by
T( f ) (x) = f (x) + (x − 1) f
′
(x)
With respect to the standard basis ϵ = {1, x, x
2
}, we have the non-diagonal matrix
A = [T]
ϵ
=
1 −1 0
0 2 −2
0 0 3
=⇒ p(t) = det(A − tI) = (1 − t)(2 − t)(3 − t)
With three distinct eigenvalues, T is diagonalizable. To find the eigenvectors, compute the
nullspaces:
λ
1
= 1: 0 = (A − λ
1
I)[v
1
]
ϵ
=
0 −1 0
0 1 −2
0 0 2
[v
1
]
ϵ
=⇒ [v
1
]
ϵ
∈ Span
1
0
0
=⇒ E
1
= Span{1}
λ
2
= 2: A − λ
2
I =
−1 −1 0
0 0 −2
0 0 1
=⇒ [v
2
]
ϵ
∈ Span
1
−1
0
=⇒ E
2
= Span{1 − x}
λ
3
= 3: A − λ
3
I =
−2 −1 0
0 −1 −2
0 0 0
=⇒ [v
3
]
ϵ
∈ Span
1
−2
1
=⇒ E
3
= Span{1 − 2x + x
2
}
Making a sensible choice of non-zero eigenvectors, we obtain an eigenbasis, with respect to
which the linear map is necessarily diagonal
β = {v
1
, v
2
, v
3
} = {1, 1 − x, 1 − 2x + x
2
} = {1, 1 − x, (1 − x)
2
}
T
a + b(1 − x) + c(1 − x)
2
= a + 2b(1 − x) + 3c(1 − x)
2
[T]
β
=
1 0 0
0 2 0
0 0 3
5
Conditions for diagonalizability of finite-dimensional operators
We now borrow a little terminology from the theory of polynomials.
Definition 1.5. Let F be a field and p(t) a polynomial with coefficients in F.
1. Let λ ∈ F be a root; p(λ) = 0. The algebraic multiplicity mult(λ) is the largest power of λ − t to
divide p(t). Otherwise said, there exists
1
some polynomial q(t) such that
p(t) = ( λ − t)
mult(λ)
q(t) and q(λ) = 0
2. We say that p(t) splits over F if it factorizes completely into linear factors; equivalently
∃a, λ
1
, . . . , λ
k
∈ F such that
p(t) = a(λ
1
− t)
m
1
· · · (λ
k
− t)
m
k
When p(t) splits, the algebraic multiplicities sum to the degree n of the polynomial
n = m
1
+ · · · + m
k
Of course, we are most interested when p(t) is the characteristic polynomial of a linear map T ∈ L(V).
If such a polynomial splits, then a = 1 and λ
1
, . . . , λ
k
are necessarily the (distinct) eigenvalues of T.
Example 1.6. The field matters! For instance p(t) = t
2
+ 1 = (t − i)(t + i) = −( i − t)(−i − t) splits
over C but not over R. Its roots are plainly ±i.
For the purposes of review, we state the main result; this will be proved in the next section.
Theorem 1.7. Let V be finite-dimensional. A linear map T ∈ L(V) is diagonalizable if and only if,
1. Its characteristic polynomial splits over F, and,
2. The geometric and algebraic multiplicities of each eigenvalue are equal; dim E
λ
j
= mult(λ
j
).
Example 1.8. The matrix A =
3 1 0
0 3 0
0 0 5
is easily seen to have eigenvalues λ
1
= 3 and λ
2
= 5. Indeed
p(t) = (3 − t)
2
(5 − t), mult(3) = 2, mult( 5) = 1
E
3
= Span
1
0
0
, E
5
= Span
0
0
1
, dim E
3
= dim E
5
= 1
This matrix is non-diagonalizable since dim E
3
= 1 = 2 = mult(3).
Everything prior to this should be review. If it feels very unfamiliar, revisit your notes from 121A,
particularly sections 5.1 and 5.2 of the textbook.
1
The existence follows from Descartes factor theorem and the division algorithm for polynomials.
6
Exercises 1.1 1. For each matrix over R; find its characteristic polynomial, its eigenvalues/spaces,
and its algebraic and geometric multiplicities; decide if it is diagonalizable.
(a) A =
2 0 0 0
0 3 1 0
0 0 3 1
0 0 0 3
(b) B =
−1 6 0 0
−2 6 0 0
0 0 3 0
0 0 0 3
2. Suppose A is a real matrix with eigenpair (λ, v). If λ ∈ R show that (λ, v) is also an eigenpair.
3. Show that the characteristic polynomial of A =
3 −4
4 3
does not split over R. Diagonalize A
over C.
4. Give an example of a 2 × 2 matrix whose entries are rational numbers and whose characteristic
polynomial splits over R, but not over Q.
5. Diagonalize L
C
∈ L(C
2
) where C =
2i 1
2 0
.
6. If p(t) splits, explain why
det T = λ
mult(λ
1
)
1
· · · λ
mult(λ
k
)
k
where λ
1
, . . . , λ
k
are the distinct eigenvalues of T.
7. Suppose T ∈ L(V) is invertible with eigenvalue λ. Prove that λ
−1
is an eigenvalue of T
−1
with
the same eigenspace E
λ
. If T is diagonalizable, prove that T
−1
is also diagonalizable.
8. If V is finite-dimensional and T ∈ L(V), we may define det T to equal det[T]
β
, where β is any
basis of V. Explain why the choice of basis does not matter; that is, if γ is any other basis of V,
we have det[T]
γ
= det[T]
β
.
7
1.2 Invariant Subspaces and the Cayley–Hamilton Theorem
The proof of Theorem 1.7 is facilitated by a new concept, of which eigenspaces are a special case.
Definition 1.9. Suppose T ∈ L(V). A subspace W of V is T-invariant if T(W) ⊆ W. In such a case,
the restriction of T to W is the linear map
T
W
: W → W : w 7→ T(w)
Examples 1.10. 1. The trivial subspace {0} and the entire vector space V are invariant for any
linear map T ∈ L(V).
2. Every eigenspace is invariant; if v ∈ E
λ
, then T(v) = λv ∈ E
λ
.
3. Continuing Example 1.8, if A =
3 1 0
0 3 0
0 0 5
then W = Span
{
i, j
}
is an invariant subspace for the
linear map L
A
. Indeed
A(xi + yj) = (3x + y)i + 3yj ∈ W
W is an example of a generalized eigenspace; we’ll study these properly at the end of term.
To prove our diagonalization criterion, we need to see how to factorize the characteristic polynomial.
It turns out that factors of p(t) correspond to T-invariant subspaces!
Example 1.11. W = Span{i, j} is an invariant subspace of A =
1 2 4
0 3 1
0 0 2
∈ M
3
(R ). With respect to
the standard basis, the restriction [L
A
]
W
has matrix
1 2
0 3
. The characteristic polynomial p
W
( t) of the
restriction is plainly a factor of the whole,
p(t) = (1 − t)(2 − t)(3 − t) = (2 − t)p
W
( t)
Theorem 1.12. Suppose T ∈ L(V), that dim V is finite and that W is a T-invariant subspace of V.
Then the characteristic polynomial of the restriction T
W
divides that of T.
The proof simply abstracts the approach of the example.
Proof. Extend a basis β
W
of W to a basis β of V. Since T(w) ∈ Span β
W
for each w ∈ W, we see that
the matrix of [T] has block form
[T]
β
=
A B
O C
=⇒ p(t) = det(A − tI) det(C − tI) = p
W
( t) det( C − tI)
where p
W
( t) is the characteristic polynomial, and A = [T
W
]
β
W
the matrix of the restriction T
W
.
Corollary 1.13. If λ is an eigenvalue of T, then T
E
λ
= λI
E
λ
is a multiple of the identity, whence,
1. The characteristic polynomial of the restriction T
E
λ
is p
λ
( t) = (λ − t)
dim E
λ
.
2. p
λ
( t) divides the characteristic polynomial of T. In particular dim E
λ
≤ mult(λ).
8
We are now in a position to state and prove an extended version of Theorem 1.7.
Theorem 1.14. Suppose dim
F
V = n and that T ∈ L(V) has distinct eigenvalues λ
1
, . . . , λ
k
. The
following are equivalent:
1. T is diagonalizable.
2. The characteristic polynomial splits over F and dim E
λ
j
= mult(λ
j
) for each j; indeed
p(t) = p
λ
1
( t) · · · p
λ
k
( t) = (λ
1
− t)
dim E
λ
1
· · · (λ
k
− t)
dim E
λ
k
3.
k
∑
j=1
dim E
λ
j
= n
4. V = E
λ
1
⊕ · · · ⊕ E
λ
k
Example 1.15. A =
7 0 −12
0 1 0
2 0 −3
is diagonalizable. Indeed p(t) = (1 − t)
2
(3 − t) splits, and we have
λ
1 3
mult(λ) 2 1
E
λ
Span
n
2
0
1
,
0
1
0
o
Span
3
0
1
dim E
λ
2 1
and R
4
= E
1
⊕ E
3
With respect to the eigenbasis β =
n
2
0
1
,
0
1
0
,
3
0
1
o
, the map is diagonal [L
A
]
β
=
1 0 0
0 1 0
0 0 3
Proof. (1 ⇒ 2) If T is diagonalizable with eigenbasis β, then [T]
β
is diagonal. But then
p(t) = ( λ
1
− t)
m
1
· · · (λ
k
− t)
m
k
splits and
∑
mult(λ
i
) = n. The cardinality n of an eigenbasis is at most
∑
dim E
λ
i
since every
element is an (independent) eigenvector. By Corollary 1.13 (dim E
λ
j
≤ mult(λ
j
)) we see that
n ≤
∑
dim E
λ
j
≤
∑
mult(λ
j
) = n =⇒ ∀j, dim E
λ
j
= mult(λ
j
)
whence the inequalities are equalities with each pair equal dim E
λ
j
= mult(λ
j
)
(2 ⇒ 3) p(t) splits =⇒ n =
∑
mult(λ
j
) =
∑
dim E
λ
j
(3 ⇒ 4) Assume E
λ
1
⊕ · · · ⊕ E
λ
j
exists.
2
If (λ
j+1
, v
j+1
) is an eigenpair, then v
j+1
∈ E
λ
1
⊕ · · · ⊕ E
λ
j
for
otherwise this would contradict Lemma 1.2.
By induction, E
λ
1
⊕ · · · ⊕ E
λ
k
exists; by assumption it has dimension n = dim V and therefore
equals V.
(4 ⇒ 1) For each j, choose a basis β
j
of E
λ
j
. Then β := β
1
∪ · · · ∪ β
k
is a basis of V consisting of
eigenvectors of T; an eigenbasis.
2
Distinct eigenspaces have trivial intersection: i
1
= i
2
≤ j =⇒ E
i
1
∩ E
i
2
= {0}.
9
T-cyclic Subspaces and the Cayley–Hamilton Theorem
We finish this chapter by introducing a general family of invariant subspaces and using them to prove
a startling result.
Definition 1.16. Let T ∈ L(V) and let v ∈ V. The T-cyclic subspace generated by v is the span
⟨
v
⟩
= Span{v, T(v), T
2
( v), . . .}
Example 1.17. Recalling Example 1.10.3 Let A =
3 1 0
0 3 0
0 0 5
, and v = i + k. It is easy to see that
Av = 3i + 5k, A
2
v = 9i + 25k, . . . , A
m
v = 3
m
i + 5
m
k
all of which lie in Span{i, k}. Plainly this is the L
A
-cyclic subspace
⟨
i + k
⟩
.
The proof of the following basic result is left as an exercise.
Lemma 1.18.
⟨
v
⟩
is the smallest T-invariant subspace of V containing v, specifically:
1.
⟨
v
⟩
is T-invariant.
2. If W ≤ V is T-invariant and v ∈ W, then
⟨
v
⟩
≤ W.
3. dim
⟨
v
⟩
= 1 ⇐⇒ v is an eigenvector of T.
We were lucky in the example that the general form A
m
v was so clear. It is helpful to develop a more
precise test for identifying the dimension and a basis of a T-cyclic subspace.
Suppose a T-cyclic subspace
⟨
v
⟩
= Span{v, T(v), T
2
( v), . . .} has finite dimension.
3
Let k ≥ 1 be
maximal such that the set
{v, T(v), . . . , T
k−1
( v)}
is linearly independent.
• If k doesn’t exist, the infinite linearly independent set {v, T(v), . . .} contradicts dim
⟨
v
⟩
< ∞.
• By the maximality of k, T
k
( v) ∈ Span{v, T(v), . . . , T
k−1
( v)}; by induction this extends to
j ≥ k =⇒ T
j
( v) ∈ Span{v, T(v), . . . , T
k−1
( v)}
It follows that
⟨
v
⟩
= Span{v, T(v), . . . , T
k−1
( v)}, and we’ve proved a useful criterion.
Theorem 1.19. Suppose v = 0, then
dim
⟨
v
⟩
= k ⇐⇒ {v, T(v), . . . , T
k−1
( v)} is a basis of
⟨
v
⟩
⇐⇒ k is maximal such that {v, T(v), . . . , T
k−1
( v)} is linearly independent
⇐⇒ k is minimal such that T
k
( v) ∈ Span{v, T(v), . . . , T
k−1
( v)}
3
Necessarily the situation if dim V < ∞, when we are thinking about characteristic polynomials.
10
Examples 1.20. 1. According to the Theorem, in Example 1.17 we need only have noticed
• v = i + k and Av = 3i + 5k are linearly independent.
• That A
2
( i + k) = 9i + 25k ∈ Span{v, Av}.
We could then conclude that
⟨
v
⟩
= Span{v, Av} has dimension 2.
2. Let T(p(x)) = 3p(x) − p
′′
(x) viewed as a linear map T ∈ L(P
2
(R )) and consider the T-cyclic
subspace generated by the polynomial p(x) = x
2
T(x
2
) = 3x
2
− 2, T
2
(x
2
) = T(3x
2
− 2) = 3(3x
2
− 2) − 6 = 9x
2
− 12, . . .
Observe that {x
2
, T(x
2
) } is linearly independent, but that
T
2
(x
2
) = 9x
2
− 12 = −9x
2
+ 6(3x
2
− 2) ∈ Span{x
2
, T(x
2
) }
We conclude that dim
x
2
= 2. An alternative basis for
x
2
is plainly {1, x
2
}.
We finish by considering the interaction of a T-cyclic subspace with the characteristic polynomial.
Surprisingly, the coefficients of the characteristic polynomial and the linear combination coincide.
Continuing the Example, if W =
x
2
and β
W
= {x
2
, T(x
2
) } = {x
2
, 3x
2
− 2}, then
[T
W
]
β
W
=
0 −9
1 6
=⇒ p
W
( t) = t
2
− 6t + 9
Theorem 1.21. Let T ∈ L(V) and suppose W =
⟨
w
⟩
has dim W = k with basis
β
W
= {w, T(w), . . . , T
k−1
( w)}
in accordance with Theorem 1.19, then
1. If T
k
( w) + a
k−1
T
k−1
( w) + · · · + a
0
w = 0, then the characteristic polynomial of T
W
is
p
W
( t) = (−1)
k
t
k
+ a
k−1
t
k−1
+ · · · + a
1
t + a
0
2. p
W
(T
W
) = 0 is the zero map on W.
Proof. 1. This is an exercise.
2. Write S ∈ L(V) for the linear map
S := p
W
(T) = (−1)
k
T
k
+ a
k−1
T
k−1
+ · · · + a
0
I
Part 1 says S(w) = 0. Since S is a polynomial in T, it commutes with all powers of T:
∀j, S(T
j
( w)) = T
j
(S( w)) = 0
Since S is zero on the basis β
W
of W, we see that S
W
is the zero function.
11
With a little sneakiness, we can drop the W’s in the second part of the Theorem and observe an
intimate relation between a linear map and its characteristic polynomial.
Corollary 1.22 (Cayley–Hamilton). If V is finite-dimensional, then T ∈ L(V) satisfies its charac-
teristic polynomial; p(T) = 0.
Proof. Let w ∈ V and consider the cyclic subspace W =
⟨
w
⟩
generated by w. By Theorem 1.12,
p(t) = q
W
( t)p
W
( t)
for some polynomial q
W
. But the previous result says that p
W
(T) (w) = 0, whence
p(T)(w) = 0
Since we may apply this reasoning to any w ∈ V, we conclude that p(T) is the zero function.
Examples 1.23. 1. A =
2 1
3 4
has p(t) = t
2
− 6t + 5 and we confirm:
A
2
− 6A =
7 6
18 19
− 6
2 1
3 4
= −5I
It may seem like a strange thing to do for this matrix, but the characteristic equation can be
used to calculate the inverse of A:
A
2
− 6A + 5I = 0 =⇒ A(A − 6I) = −5I =⇒ A
−1
=
1
5
(6I − A) =
1
5
4 −1
−3 2
2. We use the Cayley–Hamilton Theorem to compute A
4
when
A =
2 −1
8
3
0 1 −6
0 0 2
The characteristic polynomial is
p(t) = (2 − t)
2
(1 − t) = 4 − 8t + 5t
2
− t
3
By Cayley–Hamilton,
A
4
= AA
3
= A(5A
2
− 8A + 4I)
= 5A
3
− 8A
2
+ 4A = 5(5A
2
− 8A + 4I) − 8A
2
+ 4A
= 17A
2
− 36A + 20I = 17
4 −3
50
3
0 1 −18
0 0 4
− 36
2 −1
8
3
0 1 −6
0 0 2
+ 20
1 0 0
0 1 0
0 0 1
=
16 −15
562
3
0 1 −90
0 0 16
12
3. Recall Example 1.4.2, where the linear map T( f (x)) = f (x) + (x − 1) f
′
(x) had
p(t) = (1 − t)(2 − t)(3 − t) = −t
3
+ 6t
2
− 11t + 6
By Cayley–Hamilton, T
3
= 6T
2
− 11T + 6I. You can check this explicitly, after first computing
T
2
( f (x)) = f (x) + 3(x − 1) f
′
(x) + (x − 1)
2
f
′′
(x), etc.
Cayley–Hamilton can also be used to simplify higher powers of T and even to compute the
inverse!
I =
1
6
(T
3
− 6T
2
+ 11T) =⇒ T
−1
=
1
6
(T
2
− 6T + 11I)
=⇒ T
−1
( f (x)) = f (x) −
1
2
(x − 1) f
′
(x) +
1
6
(x − 1)
2
f
′′
(x)
Exercises 1.2 1. For the linear map T = L
A
: R
3
→ R
3
where A =
3 0 0
0 2 4
0 0 2
find the T-cyclic
subspace generated by the standard basis vector e
3
=
0
0
1
.
2. Let T = L
A
, where A =
1 2 4
0 3 1
0 0 2
and let v =
0
1
−1
. Compute T(v) and T
2
( v). Hence describe
the T-cyclic subspace
⟨
v
⟩
and its dimension.
3. Given A =
2 0 0 0
0 3 1 0
0 0 3 1
0 0 0 3
, find two distinct L
A
-invariant subspaces W ≤ R
4
such that dim W = 3.
4. Suppose that W and X are T-invariant subspaces of V. Prove that the sum
W + X = {w + x : w ∈ W, x ∈ X}
is also T-invariant.
5. Prove Lemma 1.18.
6. Give an example of an infinite-dimensional vector space V, a linear map T ∈ L(V), and a vector
v such that
⟨
v
⟩
= V.
7. Let β = {sin x, cos x, 2x sin x, 3x cos x} and T =
d
dx
∈ L(Span β). Plainly the subspace W :=
Span{sin x, cos x} is T-invariant. Compute the matrices [T]
β
and [T
W
]
β
W
and observe that
p(t) =
p
W
( t)
2
8. Verify explicitly that A =
2 3
0 2
satisfies its characteristic polynomial.
9. Check the details of Example 1.23.3 and evaluate T
4
as a linear combination of I, T and T
2
. In
particular, check the evaluation of T
−1
( f (x)).
13
10. Suppose a, b are constants with a = 0 and define T( f (x)) = a f (x) + b f
′
(x).
(a) Find an expression for the inverse T
−1
( f (x)) if T ∈ L(P
1
(R ))
(b) Find an expression for the inverse T
−1
( f (x)) if T ∈ L(P
2
(R ))
Your answers should be written in terms of f and its derivatives.
11. Let T( f ) (x) = f
′
(x) +
1
x
R
x
0
f (t) dt be a linear map T ∈ L(P
2
(R )).
(a) Find the characteristic polynomial of T and identify its eigenspaces. Is T diagonalizable?
(b) Find a, b, c ∈ R such that T
3
= aT
2
+ bT + cI.
(c) What are dim L(P
2
(R )) and dim Span{T
k
: k ∈ N
0
}? Explain.
12. If A =
a b
c d
has non-zero determinant, use the Cayley–Hamilton Theorem to obtain the usual
expression for A
−1
.
13. Recall Examples 1.10.3, 1.17, and 1.20.1 with A =
3 1 0
0 3 0
0 0 5
.
(a) If v =
x
y
z
, show that det(v, Av, A
2
v) = −4y
2
z
(b) Hence determine all L
A
-cyclic subspaces of R
3
.
14. (a) Consider Example 1.20.2 where T ∈ L(P
2
(R )) is defined by T(p(x)) = 3p(x) − p
′′
(x).
Prove that all T-cyclic subspaces have dimension ≤ 2.
(b) What if we instead consider S ∈ L(P
2
(R ) defined by S(p(x)) = 3p(x) − p
′
(x)?
15. We prove part 1 of Theorem 1.21.
(a) Explain why the matrix of T
W
with respect to the basis β
W
is
[T
W
]
β
W
=
0 0 0 · · · 0 −a
0
1 0 0 0 −a
1
0 1 0 0 −a
2
.
.
.
.
.
.
.
.
.
0 0 0 0 −a
k−2
0 0 0 · · · 1 −a
k−1
∈ M
k
(F )
(b) Compute the characteristic polynomial p
W
( t) = det
[T
W
]
β
W
− tI
k
by expanding the de-
terminant along the first row.
14
