Problem Paper 2
th
Due October 27 , 2015 (no later than 3.30 pm - secretary room)
Problem 1 (10 MARKS)
Prove the Steinitz exchange lemma that is, let (V, K) be a vector space and
consider the following linearly independent set of vectors in V , namely
S = {v 1 , · · · , v m },
W = {w1 , · · · , wn }
with m < n. Show that there exists a vector wj ∈ W for some j = 1, · · · , n
such that the set S ∪ {wj } is linearly independent. Hint: the following
lemma could be useful in proving this result.
Lemma 1 Let A ∈ M(m,n) (K) and x
Ax = 0 with m < n where

a11 a12 · · ·
 a21 a22 · · ·

A =  ..
..
..
 .
.
.
am1 am2 · · ·
∈ Kn . Then, the homogeneous system
a1n
a2n
..
.



,




x=

amn
x1
x2
..
.





xn
has at least one non-trivial solution, that is not all acomponents of x vanish.
Problem 2 (10 MARKS)
In this exercise you will prove some consequences of the Steinitz exchange
lemma. Let (V, K) be a finite-dimensional vector space of dimension n ∈ N.
Prove the following statements. Each proof is 5 marks worth.
′
1. Let # denote the cardinality of a set. If B and B are two bases for V ,
′
then #B = #B = n.
2. Any linearly independent set {v 1 , · · · , v s } with s ≤ n can be extended
to a basis for V .
Problem 3 (10 MARKS)
Let us recall that two vector spaces (V, K) and (W, K) are isomorphic if
there exists an isomorphism, that is a linear and bijective map φ : V −→ W .
Show that every finite-dimensional vector space (V, K) with dim(V ) = n is
1
isomorphic to the vector space (Kn , K). Hint: show that the map φ : Kn −→
V such that it associates to the n-tuple (α1 , · · · , αn ) ∈ Kn the vector
φ(v) = α1 v 1 + · · · + αn v n
with B = {v 1 , · · · , v n } basis for V is linear and bijective.
Problem 4 (5 MARKS)
Consider the vector space (R2 , R). Show that the subset
U = {(x, y) ∈ R2 | y = mx, m ∈ R}
is a subspace of V .
Problem 5 (5 MARKS)
Consider the vector space (V, K) and a collection U = {Uλ | λ ∈ I} of
subspaces of V . Here, I denotes an indexing set that could be finite such as
I = {1, 2, · · · , n} with n ∈ N or numerable such as I = N or I = Z. Show
that the intersection of any collection of subspaces is again a subspace.
Problem 6 (10 MARKS)
Let (V, K) be a finite-dimensional vector space and U, W finite-dimensional
subspaces of V . Show that
dim(U + W ) = dim(U ) + dim(W ) − dim(U ∩ W ).
Solution
We need to consider the following cases.
1. If U ⊆ W then U +W = W and U ∩W = U and the formula is trivially
true since
dim(W ) = dim(U )+dim(W )−dim(U ) = dim(U )+dim(W )−dim(U ∩W ).
Similar treatment for W ⊆ U .
2. If U ∩ W = {0}, let BU and BW be bases for U and W , respectively.
Then, U + W = span (BU ∪ BW ) and BU ∪ BW is a basis for U + W .
Then,
dim(U + W ) = #BU + #BW = dim(U ) + dim(W ).
2
3. Let U ∩ W ̸= {0} and not coinciding with U or W . Since U and W are
finite-dimensional subspaces, then U ∩ W is again a finite-dimensional
vector space with basis BU ∩W = {z 1 , · · · , z r }. Since U ∩ W ⊂ U and
U ∩ W ⊂ W , Steinitz exchange lemma allows to extend BU ∩W to bases
for U and W as follows
BU = BU ∩W ∪ {u1 , · · · , um }, u1 , · · · , um ∈ U,
BW = BU ∩W ∪ {w1 , · · · , wn }, w1 , · · · , wn ∈ W.
Hence,
dim(U ∩ W ) = r,
dim(U ) = r + m,
dim(W ) = r + n.
By definition of sum of subspaces
U + W = span(BU ∪ BW ),
= span(BU ∩W ∪ {u1 , · · · , um } ∪ BU ∩W ∪ {w1 , · · · , wn }),
= span(BU ∩W ∪ {u1 , · · · , um } ∪ {w1 , · · · , wn })
and therefore
dim(U + W ) = r + m + n = r + m + n + r − r = (r + m) + (n + r) − r,
= dim(U ) + dim(W ) − dim(U ∩ W )
and the proof is completed. 3