Chapter 2: Finite-Dimensional Vector Spaces

Linear Algebra Done Right (4th Edition), by Sheldon Axler

Last updated: November 1, 2024

Contents
2A: Span and Linear Independence 2
2A Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2B: Bases 6
2B Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2C: Dimension 9
2C Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1
2A: Span and Linear Independence
Definition 1 (Linear Combination). A linear combination of a list v1 , . . . , vm
of vectors in V is a vector of the form
a1 v1 + · · · + am vm
where a1 , . . . , am ∈ F.
Definition 2 (Span). The set of all linear combinations of list of vectors
v1 , . . . , vm in V is called the span of v1 , . . . , vm , denoted by span(v1 , . . . , vm ). In
other words,
span(v1 , . . . , vm ) = {a1 v1 + · · · am vm : a1 , . . . , am ∈ F}
The span of the empty list () is defined to be {0}.
Theorem 3. The span of a list of vectors in V is the smallest subspace of V
containing all vectors in the list.
Definition 4 (Spans). If span(v1 , . . . , vm ) equals V , we say the list v1 , . . . , vm
spans V .
Definition 5 (Finite-dimensional vector space). A vector space is called finite-
dimensional if some list of vectors in it spans the space.
Definition 6 (polynomial). A function p : F → F is called a polynomial with
coefficients in F if there exist a0 , . . . , am ∈ F s.t.
(z) = a0 + a1 z + a2 z 2 + · · · + am z m
for all z ∈ F.
P(F) is the set of all polynomials with coefficients in F.
Definition 7 (Linear independence). A list of vectors v1 , . . . , vm in V is called
linearly independent if the only choice of a1 , . . . , am ∈ F that makes
a1 v1 + · · · + am vm = 0
is a1 = · · · = am = 0.
The empty list () is also declared to be linearly independent.
Lemma 8. Suppose v1 , . . . vm is a linearly dependent list in V . Then there
exists k ∈ {1, 2, . . . , m} s.t.
vk ∈ span(v1 , . . . , vk−1 )
Furthermore, if k satisfies the condition above and the k th term is removed from
v1 , . . . vm , then the span of the remaining list equals span(v1 , . . . , vm ).
Lemma 9 (length of linearly independent list ≤ length of spanning list). In a
finite-dimensional vector space, the length of every linearly independent list of
vectors is less than or equal to the length of every spanning list of vectors.

2
 Problem 1
Find a list of four distinct vectors in F3 whose span equals

{(x, y, z) ∈ F3 : x + y + z = 0}

Proof. Example: (1, 0, −1), (0, −1, 1), (1, −1, 0), (1, −2, 1).

Problem 2
Prove or give a counterexample: If v1 , v2 , v3 , v4 spans V , then the list

v1 − v2 , v2 − v3 , v3 − v4 , v4

also spans V .

Proof. Take any v ∈ V , then we have v = a1 v1 + a2 v2 + a3 v3 + a4 v4 = a1 (v1 −

v2 )+(a1 +a2 )(v2 −v3 )+(a1 +a2 +a3 )(v3 −v4 )+(a1 +a2 +a3 +a4 )v4 . Conversely,
any linear combination of these vectors still belong to V .

Problem 3
Suppose v1 , . . . , vm is a list of vectors in V . For k ∈ {1, . . . , m}, let

wk = v1 + · · · + vk

Show that span(v1 , . . . , vm ) = span(w1 , . . . , wm )

Pm
Proof. Take v = i=1 ai vi from l.h.s, then we can write

v = a1 v1 + · · · am vm
= (a1 − am )v1 + · · · (am−1 − am )vm−1 + am wm
= (a1 − am − am−1 )v1 + · · · (am−2 − am−1 − am )vm−2 + (am−1 − am )wm−1 + am wm
 
Xm m
X
= ai − aj  wi ∈ r.h.s
i=1 j=i+1

Pm Pm Pi
Conversely, take w = i=1 bi wi = i=1 (bi j=1 cj )vj ∈ l.h.s..

Problem 4
(a) Show that a list of length one in a vector space is linearly independent
if and only if the vector in the list is not 0.
(b) Show that a list of length two in a vector space is linearly independent
if and only if neither of the two vectors in the list is a scalar multiple of
the other.

3
Proof. (a) In order for the only way to write av = 0 is to ensure v ̸= 0.
(b) av1 + bv2 = 0. ⇒ the only solution is a = b = 0 so v1 cannot be a multiple
of v2 ; vice versa. ⇐ same reason.

Problem 8
Suppose v1 , v2 , v3 , v4 is linearly independent in V . Prove that the list

v1 − v2 , v2 − v3 , v3 − v4 , v4

is also linearly independent.

Proof.

a1 (v1 −v2 )+a2 (v2 −v3 )+a3 (v3 −v4 )+a4 v4 = a1 v1 +(−a1 +a2 )v2 +(−a2 +a3 )v3 +(−a3 +a4 )v4

The only solution is a1 = 0, −a1 + a2 = 0, −a2 + a3 = 0, −a3 + a4 = 0.

Problem 9
Prove or give a counter example: If v1 , . . . , vm is a linearly independent
list of vectors in V , then

5v1 − 4v2 , v2 , v3 , . . . , vm

is also linearly independent.

Proof.

a1 (5v1 − 4v2 ) + a2 v2 + . . . am vm
= 5a1 v1 + (−4a1 + a2 )v2 + . . . am vm

The only solution is 5a1 = −4a1 + a2 = . . . = am = 0.

Problem 10
Prove or give a counterexample: If v1 , . . . , vm is a linearly independent
list of vectors in V and λ ∈ F with λ =
̸ 0, then λv1 , . . . , λvm is linearly
independent.

Proof.
m
X
λai vi = 0
i

The only solution is ai = 0 for all i.

Problem 12
Suppose v1 , . . . , vm is linearly independent in V and w ∈ V . Prove that
if v1 + w, . . . , vm + w is linearly dependent, then w ∈ span(v1 , . . . , vm ).

4
 Pm
Proof. We know the only solution to ai vi is all ai = 0. P
Now we have
Pm Pim m
a v
that non-zero ai for solving i ai vi + i ai w = 0. Thus w = Pi m ai i i which
i
completes the proof.

Problem 13
Suppose v1 , . . . , vm is linearly independent in V and w ∈ V . Show that

v1 , . . . , vm , w is linearly independent ⇐⇒ w ∈
/ span(v1 , . . . , vm )

Proof. ⇒ By contradiction, if w in the span then it can be written as linear

combination for some nonzero ai and thus they are linearly dependent.
⇐ Similarly.

Problem 17
Prove that V is infinite-dimensional if and only if there is a sequence
v1 , v2 , . . . of vectors in V such that v1 , . . . , vm is linearly independent for
every positive integer m.

Proof. ⇒ There doesn’t exist any finite list of vectors that span the space. For
the sake of contradicting assumes for every sequence v1 , . . . of vectors in V ∃ m
such that v1 , . . . , vm is linearly dependent. Then this means we can construct
the basis for the vector space as follows: select vi ∈ V s.t. vi and v1 , . . . , vi−1
are linearly independent. This means that there exits m s.t. v1 , . . . , vm that
spans V , forming a contradiction.
⇐ Suppose for contradiction. Then there exists a span, contradicting the
linear independence claim for every m.

Problem 18
Prove F∞ is infinite-dimensional.

Proof. We apply Problem 17. Construct vi to be the vector that has 1 on the
i-th coordinate and 0 elsewhere. clearly v1 , v2 , . . . are s.t. v1 , . . . , vm is linearly
independent for every positive integer m.

5
2B: Bases
Definition 10 (basis). A basis of V is a list of vectors in V that is linearly
independent and spans V .
Theorem 11 (criterion for basis). A list v1 , . . . , vm of vectors in V is a basis
of V if and only if every v ∈ V can be written uniquely in the form
m
X
v= ai vi
i

where ai ∈ F.
Lemma 12 (every spanning list contains a basis). Every spanning list in a
vector space can be reduced to a basis of the vector space.
Lemma 13. Every finite-dimensional vector space has a basis.

Lemma 14. Every linearly independent list of vectors in a finite-dimensional

vector space can be extended to a basis of the vector space.
Lemma 15. Suppose V is finite-dimensional and U is a subspace of V . Then
there is a subspace W of V such that V = U ⊕ W .

6
 Problem 1
Find all vector spaces that have exactly one basis.

Proof. The only answer is {0}. Otherwise, for any basis v one can get av for
a ̸= 0, a ̸= 1.

Problem 4
1. Let U be the subspace of C5 defined by U = {(z1 , z2 , z3 , z4 , z5 ) ∈
C5 : 6z1 = z2 , z3 + 2z4 + 3z5 = 0} Find a basis of U.
2. Extend the basis to a basis in C5 .

3. Find a subspace W of C5 s.t. C5 = U ⊕ W.

Proof. 1. (z1 , 6z1 , −2z4 − 3z5 , z4 , z5 ) ⇒ {(1,6,0,0,0), (0,0,-2,1,0), (0,0,-3,0,1)}

2. {(1, 6, 0, 0, 0), (0, 0, −2, 1, 0), (0, 0, −3, 0, 1), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0)}
3. W = span({(0, 1, 0, 0, 0), (0, 0, 1, 0, 0)})

Problem 5
Suppose V is finite-dimensional and U, W are subspaces of V such that
V = U + W. Prove that there exists a basis of V consisting of vectors in
U ∪ W.

Proof. Let {vi }m

i=1 denote the basis for the vector space V . By definition we
have vi =Pui + wi for some ui , wi . Then we have the spanning set of the vector
m
space V i ai (ui + wi ), which can be reduced to a basis by the lemma.

Problem 7
Suppose v1 , v2 , v3 , v4 is a basis of V . Prove that

v1 + v2 , v2 + v3 , v3 + v4 , v4

is also a basis of V .

Proof. We know v1 , v2 , v3 , v4 is linearly independent and spans V .

a1 (v1 +v2 )+a2 (v2 +v3 )+a3 (v3 +v4 )+a4 v4 = a1 v1 +(a1 +a2 )v2 +(a2 +a3 )v3 +(a3 +a4 )v4
which shows the linear independence. For proving spanning, let v ∈ V then

4
X
v= ai vi = a1 (v1 + v2 ) + (a2 − a1 )(v2 + v3 ) + (a3 − a2 )(v3 + v4 ) + (a4 − a3 )v4
i=1

7
 Problem 8
Prove or give a counterexample: If v1 , v2 , v3 , v4 is a basis of V and U is a
subspace of V such that v1 , v2 ∈ U and v3 ∈ / U and v4 ∈ / U, then v1 , v2 is
a basis of U.

Proof. Take V = R4 and the standard basis. Consider U = {(x1 , x2 , x3 , kx3 )},
then we disprove the claim.

Problem 10
Suppose U and W are subspaces of V s.t. V = U ⊕ W. Suppose also that
u1 , . . . um is a basis of U and w1 , . . . , wn is a basis of W. Prove that

u1 , . . . , um , w1 , . . . , wn

is a basis of V .

Proof. We know that this set is linearly independent (otherwise violating the
direct sum assumption)
Pm soPit sufficies to prove the spanning. Let v ∈ V , then
n
v = u + w = i=1 ai ui + j=1 bj wj .

8
2C: Dimension
Lemma 16 (basis length does not depend on basis). Any two bases of a finite-
dimensional vector space have the same length.
Definition 17 (dimension). • The dimension of a finite-dimensional vector
space is the length of any basis of the vector space.
• The dimension of a finite-dimensional vector space V is denoted by dim V .

Corollary 18. If V is finite-dimensional and U is a subspace of V , then

dim U ≤ dim V .
Corollary 19. Suppose V is finite-dimensional. Then every linearly independent
list of vectors in V of length dim V is a basis of V .

Corollary 20. Suppose that V is finite-dimensional and U is a subspace of V

such that dim U = dim V . Then U = V .
Corollary 21. Suppose that V is finite-dimensional. Then every spanning list
of vectors in V of length dim V is a basis of V .

Theorem 22 (dimension of a sum). If V1 and V2 are subspaces of a finite-

dimensional vector space, then

dim(V1 + V2 ) = dim V1 + dim V2 − dim(V1 ∩ V2 )

9
 Problem 1
Show that the subspaces of R2 are precisely {0}, all lines in R2 containing
the origin and R2 .

Proof. We know dim(R2 ) = 2 so the subspace dimension is either 0 ({0}) or 1

(then this means it has to be lines crossing the origin).

Problem 5
(a) Let U = {p ∈ P4 (F) : p(2) = p(5)}. Find a basis of U.
(b) Extend the basis in (a) to a basis of P4 (F).
(c) Find a subspace W of P4 (F) s.t. P4 (F) = U ⊕ W.

Proof. (a) This means that

a0 + 2a1 + 4a2 + 8a3 + 16a4 = a0 + 5a1 + 25a2 + 125a3 + 625a4

Solving this gives that a1 = −7a2 − 39a3 − 203a4 . So we can write that

p(x) = a0 + (−7a2 − 39a3 − 203a4 )x + a2 x2 + a3 x3 + a4 x4

= a0 + a2 (x2 − 7x) + a3 (x3 − 39x) + a4 (x4 − 203x)

The basis now becomes {1, x2 − 7x, x3 − 39x, x4 − 203x}

(b) {1, x, x2 − 7x, x3 − 39x, x4 − 203x}
(c) W = {x}

Problem 8
Suppose v1 , . . . , vm is linearly independent in V and w ∈ V . Prove that

dim span(v1 + w, . . . , vm + w) ≥ m − 1

Proof. We claim that

vi+1 − vi ∈ span(v1 + w, . . . , vm + w) for all m ≥ i ≥ 2

as vi+1 − vi = (vi+1 + w) − (vi + w). Thus v2 − v1 , . . . , vm − vm−1 is in the

span(v1 + w, . . . , vm + w). We’ve proved that v2 − v1 , . . . , vm − vm−1 is linearly
independent and this list has m − 1 vectors and thus we’ve proved the claim.

Problem 9
Suppose m is a positive integer and p0 , p1 , . . . , pm ∈ P(F) are such that
each pk has degree k. Prove that p0 , p1 , . . . , pm is a basis of Pm (F).

10
Proof. It’s easy to see that this list is linearly independent. Take any element in
Pm (F), we can decompose that by degrees and get each component to be some
multiple of p′i s.

Problem 10
Suppose m is a positive integer. For 0 ≤ k ≤ m, let

pk (x) = xk (1 − x)m−k

Show that p0 , . . . , pm is a basis of Pm (F).

Proof. It suffices to prove that pk (x) are linearly independent. We know from
binomial theorem that
m  
X m i m−i
(x + y)m = xy
i=0
i

Applying this identity here we get that

m−k
X  m−k
k m − k m−i X
pk (x) = x 1 (−1)i xi = ai xk+i
i=0
i i=0

m−k


where ai =
Pm i (−1)i . For the polynomial to be identically 0
( k=0 ck pk (x) = 0), for each power xi we need the coefficient to be 0:
min(i,m)  
X m−k
ck (−1)i−k = 0
i−k
k=0

We can prove the only solution for this is all ck = 0 by induction on m.

The base case is trivial. Assume the statement holds for = m − 1, then
Pkm−1
we try to prove for pm (x), where we have P (x) − cm pm (x) = k=0 ck pk (x) = 0.
This means that cm = 0 and we’ve proved the linear independence.

Problem 11
Suppose U and W are both four-dimensional subspaces of C6 . Prove that
there exist two vectors in U ∩ W such that neither of these vectors is a
scalar multiple of the other.

Proof. dim(U ∩ W) = dim(U) + dim(W) − dim(U + W) ≥ 8 − 6 = 2

So there must exist two linearly independent vectors in the intersection.

Problem 12
Suppose that U and W are subspaces of R8 such that dim U = 3, dim W =
5 and U + W = R8 . Prove that R8 = U ⊕ W.

11
Proof. Similar to problem 11, we can get that dim(U ∩ W) = 0.

Problem 14
Suppose V is a ten-dimensional vector space and V1 , V2 , V3 are subspaces
of V with dim V1 = dim V2 = dim V3 = 7. Prove that V1 ∩ V2 ∩ V3 ̸= {0}.

Proof.
dim((V1 ∩ V2 ) + V3 ) = dim(V1 ∩ V2 ) + dim V3 − dim(V1 ∩ V2 ∩ V3 )
and also that
dim(V1 + V2 ) = dim V1 + dim V2 − dim(V1 ∩ V2 )
This gives that

dim(V1 ∩V2 ∩V3 ) = dim V1 +dim V2 +dim V3 −dim(V1 +V2 )−dim((V1 ∩V2 )+V3 )
Note that from question we know dim V1 + dim V2 + dim V3 = 21 > 20 =
2 dim V
Hence we know

dim(V1 ∩V2 ∩V3 ) > (dim(V )−dim(V1 +V2 ))+(dim(V )−dim((V1 ∩V2 )+V3 )) > 0
We’ve thus proved the claim.
Problem 16
Suppose V is finite-dimensional and U is a subspace of V with U =
̸ V.
Let n = dim V and m = dim U. Prove that there exist n − m subspaces
of V , each of dimension n − 1, whose intersection equals U.

Proof. To show existence, we can start with the basis for U : u1 , . . . , um . We

extend the basis to V : {u1 , . . . , um , v1 , . . . , T
vn−m } := K. Construct the subspace
Vi = span{K \ {vi }}. Then we know that i Vi = span{{u1 , . . . , um }}.

Problem 17
Suppose that V1 , . . . , Vm are finite-dimensional subspaces of V . Prove
that V1 + · · · + Vm is finite-dimensional and

dim(V1 + · · · + Vm ) ≤ dim V1 + · · · + dim Vm

Proof. We prove by induction on m. The base case is trivial. Assume the

statement hold for k, then for k + 1, we have that (Denote V1 + . . . + Vk = Mk )

dim(Mk + Vk+1 ) ≤ dim(V1 ) + · · · + dim(Vk+1 )

which is finite.

12
 Problem 18
Suppose V is finite-dimensional with dim V = n ≥ 1. Prove that there
exist one-dimensional subspaces V1 , . . . , Vn of V such that

V = V1 ⊕ · · · Vn

Proof. We know there are basis {v1 , . . . , vn } for V . Hence we can construct

Vi = span{vi }

Problem 19
Prove or give a counter example:

dim(V1 + V2 + V3 ) = dim V1 + dim V2 + dim V3

− dim(V1 ∩ V2 ) − dim(V1 ∩ V3 ) − dim(V2 ∩ V3 )
+ dim(V1 ∩ V2 ∩ V3 )

Proof. We know that

dim((V1 + V2 ) + V3 ) = dim(V1 + V2 ) + dim(V3 ) − dim((V1 + V2 ) ∩ V3 )

= dim(V1 ) + dim(V2 ) + dim(V3 ) − dim(V1 ∩ V2 ) − dim((V1 + V2 ) ∩ V3 )

Here we can get that

dim((V1 + V2 ) ∩ V3 ) = dim((V1 ∩ V3 ) + (V2 ∩ V3 ))

= dim(V1 ∩ V3 ) + dim(V2 ∩ V3 ) − dim(V1 ∩ V2 ∩ V3 )

By substituting this back, we can get the desired solution.

13