Solutions to the Mock Midterm Exam – Linear Algebra

Math 110, Spring 2024. Instructor: E. Frenkel

Problem 1. Let V be the subspace of R3 defined by the equation

a1 + 2a2 + 3a3 = 0.
Find a basis of V and give a proof that it is indeed a basis.

Solution. We claim that     

 2 3 
β = −1 ,  0 
 0 −1 
is a basis (note: of course, it’s just one of many possibilities). Observe that dim(V  ) = 2.
a1
Indeed, V = N (T ), where T : R3 → R is the linear transformation sending a2  to
a3
 
a
a1 + 2a2 + 3a3 . Since T sends 0 to a, we obtain that R(T ) = R. Therefore, by
0
fundamental theorem, dim(V ) = dim(R3 ) − dim(R) = 2. Since β consists of two elements,
in order to prove that β is a basis of V , it is sufficient to prove that β is linearly independent.
Two vectors are linearly independent
 if and only if they are not proportional to each other.
2
Clearly, any scalar multiple of −1 has 0 as the third entry, whereas any scalar multiple

0
 
3
of  0  has 0 as the second entry. Hence the two vectors in β are not proportional to
−1
each other.

Problem 2. Let T be the linear map P3 (C) → P3 (C) given by the formula T (p(t)) =
p(t + 1), for every p(t) ∈ P3 (C).
Compute the matrix M(T ) with respect to the standard basis of P3 (C) consisting of
monomials in the variable t.

Solution. By definition, the ith column of [T ]β is the coordinate vector of T (ti ) = (t + 1)i
with respect to β. Hence  
1 1 1 1
0 1 2 3
M(T ) = 
0 0 1 3

0 0 0 1
1
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Problem 3. Prove that a vector space V over a field F is isomorphic to F n (where n is a

positive integer) if and only if dim(V ) = n.

Solution. Since this is an “if and only if” statement, we need to prove it in both directions.
Let us first prove that if dim(V ) = n, then V is isomorphic to F n . Choose a basis β =
{x1 , . . . , xn } of V . Then, since β spans V , every vector in V can be written in the form
n
X
v= ai xi , ai ∈ F.
i=1

Moreover, since β is linearly independent, the scalars ai are uniquely defined for each v.
Define a map φβ : V → F n by the formula
a1
 

φβ (v) = .. 
 .
an
It is a linear map because φ(cv + w)β = cφβ (v) + φβ (w) for any c ∈ F . Let us show that
φβ is invertible. Define the linear transformation ψβ : F n → V by the formula
a1
 
n
.
.
ψβ ( . ) =
 
X
ai x i .
an i=1

Then it follows from the definitions of φβ and ψβ that φβ ◦ ψβ = IF n and ψβ ◦ φβ = IV .

Hence φβ is an isomorphism.
Conversely, suppose that V and F n are isomorphic. Then there is an isomorphism
ψ : F n → V . Let {e1 , . . . , en } be the canonical basis of F n and set xi = ψ(ei ). Then
a1
 
n
.
.
ψ( . ) =
 
X
ai x i .
an i=1

Since ψ is onto, the set β = {x1 , . . . , xn } generates V . Let us show that β is a linearly
independent subset of V . Suppose that
n
X
ai xi = 0.
i=1

a1 a1
   

Since the left hand side is equal to ψ( .. ), it follows that  ...  ∈ N (ψ). Since ψ is
.
an an
one-to-one, it follows that ai = 0 for all i, and so β is indeed a linearly independent subset
of V . Therefore β is a basis of V , and dim(V ) = n.
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Problem 4. Let V be a two-dimensional vector space over R and T : V → V a linear

map. Suppose that β = {x1 , x2 } and β 0 = {y1 , y2 } are two bases in V such that
x1 = y1 − y2 , x2 = 2y1 − y2 .
Find the matrix M(T ) with respect to β if M(T ) with respect to β 0 is
 
3 −2
[T ]β 0 =
−1 2

Solution. According to the formula proved in the book,

M(T )β = C −1 M(T )β 0 C,
where  

1 2
C = [x1 ]β 0 [x2 ]β 0 =
−1 −1
Therefore we find
     
−1 −2 3 −2 1 2 1 0
M(T )β = =
1 1 −1 2 −1 −1 2 4

Problem 5. Let V be an n-dimensional vector space over F and T : V → V a linear map.

Suppose that W is a k-dimensional subspace of V , which is T -invariant (that is, ∀ v ∈ W ,
we have T (v) ∈ W ). Prove that there is a basis β of V such that each of the first k columns
of the matrix M(T ) with respect to β has the following property: its last (n − k) entries
are all equal to 0.

Solution. Choose a basis {x1 , . . . , xk } of W . By a theorem from the book, we can extend
it to a basis β = {x1 , . . . , xn } of V . In M(T ), the ith column is [T (xi )]β . If i = 1, . . . , k, then
T (xi ) ∈ W because W is T -invariant. But then T (xi ), i = 1, . . . , k, is a linear combination
of x1 , . . . , xk only, which is equivalent to saying that in [T (xi )]β , where i = 1, . . . , k, the
last (n − k) entries are all equal to 0. But these [T (xi )]β , i = 1, . . . , k, are exactly the first
k columns of the matrix M(T ), so the desired statement is proved.

Problem 6. Define linear functionals f1 : P1 (R) → R and f2 : P1 (R) → R by the formulas

f1 (p(t)) = p(3), f2 (p(t)) = p(−1)
for all p(t) ∈ P1 (R).
Find the basis of P1 (R) for which {f1 , f2 } is the dual basis.

Solution. By definition, the sought-after basis consists of the polynomials p1 (t) and p2 (t)
such that fi (pj (t)) = δi,j .
Writing p1 (t) = a1 + b1 t, we obtain
a1 + 3b1 = 1,
a1 − b1 = 0.
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Solving this system, we get a1 = 1/4, b1 = 1/4.

Writing p2 (t) = a2 + b2 t, we obtain
a2 + 3b2 = 0,
a2 − b2 = 1.
Solving this system, we get a2 = 3/4, b2 = −1/4.
Hence the sought-after basis is {1/4 + 1/4t, 3/4 − 1/4t}.

Problem 7. Let V be an n-dimensional vector space over a field F and V 0 the dual space.
Given a subspace W of V , let W 0 be the subspace of V 0 which consists of all
f :V →F such that f (v) = 0, ∀ v ∈ W.
Prove that dim W 0 = n − dim W .

First solution. Choose a basis {x1 , . . . , xk } of W . By Replacement Theorem, we can

extend it to a basis {x1 , . . . , xn } of V . By definition, f ∈ W 0 if and only if
k
!
X
f ai xi = 0, ∀ai ∈ F, i = 1, . . . , k.
i=1
Since f is linear, this property is equivalent to
f (xi ) = 0, i = 1, . . . , k.
Let {f1 , . . . , fn } be the basis of V 0 which is dual to {x1 , . . . , xn }. Then any f ∈ V 0 can be
written as
Xn
f= bi f i , bi ∈ F.
i=1
Since fi (xj ) = δi,j , we find that
f (xi ) = bi .
Therefore f ∈ W if and only if bi = 0 for all i = 1, . . . , k. This means that for any f ∈ W 0 ,
0

we have
X n
f= bi f i , bi ∈ F.
i=k+1
Hence {fk+1 , . . . , fn } is a basis of W , and so dim(W 0 ) = n − k = dim(V ) − dim(W ).
0

Second solution. Choose a basis {x1 , . . . , xk } of W . Define a map T : V 0 → F k by the

formula
f (x1 )
 

f 7→  ... 
f (xn )
0
Since each f ∈ V is a linear functional, T is a linear transformation. Let us show that
N (T ) = W 0 . Indeed, f ∈ N (T ) if and only if f (xi ) = 0 for all i = 1, . . . , k. This is
equivalent to f (v) = 0 for all v, as explained in the First Solution above.
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Let us show that R(T ) = F k (i.e., T is onto). By a theorem from the book, we can extend
the basis {x1 , . . . , xk } to a basis {x1 , . . . , xn } of V . Let {f1 , . . . , fn } be the dual basis of
V 0 . Given any n-tuple a = (a1 , . . . , an ), where ai ∈ F , define fa ∈ V 0 by the formula
X n
fa = ai f i .
i=1

a1
 

Then such that fa (xi ) = ai . If we take a = (a1 , . . . , ak , 0, . . . , 0), then T (fa ) = ...  ∈ F k ,

ak
so T is indeed onto.
Applying the fundamental theorem to T , we obtain: dim(W 0 ) = n − k.

Third solution. Let S : W → V be the linear transformation defined by the formula

S(v) = v (S is called “inclusion” or “embedding”). Its dual S 0 is a linear map S 0 : V 0 → W 0 .
We claim that N (S 0 ) is W 0 . Indeed, f ∈ N (S 0 ) if and only if f (v) = 0 for all v ∈ W , which
is precisely that condition f has to satisfy to be in W 0 . Next, we claim that R(S 0 ) = W 0
(i.e., S 0 is onto). For that, choose a basis {x1 , . . . , xk } of W and its extension to a basis
{x1 , . . . , xn } of V , as in the First Solution above. Given any linear functional f : W → F ,
we can extend it, using the dual basis of V 0 , to a linear functional fe : V → F such that
fe(xi ) = f (xi ) for all i = 1, . . . , k and fe(xi ) = 0 for all i = k + 1, . . . , n. But then S 0 (fe) = f ,
so S 0 is indeed onto.
Applying the fundamental theorem to S 0 , we obtain: dim(W 0 ) + dim(W 0 ) = dim(V 0 ).
Since dim(V 0 ) = dim(V ) and dim(W 0 ) = dim(W ), this completes the proof.