Math 115A Homework 3: Linear Transformations and Matrix Representations | Assignments Linear Algebra

Math 115A Homework 3 Due November 7th, 2008

1. Let Id :R3→R3be the identity linear transformation. For the ordered bases A,B,

compute the matrix representation [Id]B

A. (To be clear, the bases are ordered by reading

from left to right.) (2 pts each)

a) A={2e2, e3−e1, e1}and B={e1,2e2, e3−e1}.

b) A={e1, e1+e2, e1+e2+e3}and B={2e1−3e2, e3+e2,4e2}.

2. Prove: If Fis a field, Vis an F-vector space of dimension n, and Ais any ordered

basis of V, then the matrix representation [Id]A

Aof the identity linear transformation is the

n×n-identity matrix In.(4 pts)

This is essentially obvious. Let A={v1, . . . , vn}. The i-th column of the matrix [Id]A

Ais the

column vector [Id(vi)]A=ei. That is precisely the i-th column of the identity matrix.

3. Prove: Let Fbe a field and let Vbe an F-vector space of dimension n. Suppose T:V→V

is a linear transformation of rank n. Then there are ordered bases Aand Bof Vsuch that

[T]B

A=Inis the n×n-identity matrix. (4 pts)

Choose any ordered basis A={v1, . . . , vn}. The set {T(v1), . . . , T (vn)}spans the range of

T, which is V; since Vhas dimension n,B={T(v1), . . . , T (vn)}is an ordered basis of V.

The i-th column of [T]B

Ais [T(vi)]B=ei, hence [T]B

A=In.

4. Let Fbe a field and let Vand Wbe F-vector spaces. Suppose v∈V. Prove that the

map Ev:L(V, W )→Wdefined by Ev(T) = T(v) is a linear transformation. Moreover,

show that Evis onto provided v6= 0. (You may assume that Vhas finite dimension.) (4 pts)

Let Sand Tbe linear transformations from Vto Wand let λ∈Fbe a scalar. Then

Ev(S+λT )=(S+λT )(v) = S(v) + λT (v) = Ev(S) + λEv(T)

where the first equality is by definition of the addition and scalar multiplication on L(V, W ).

That is, Evis linear.

Now suppose v6= 0. Suppose w∈W. Let S⊆Vbe any basis of V. By the replacement

theorem, we can find a subset T⊆Ssuch that B:= T∪ {v}is a basis of V. Now there is a

unique linear transformation T:V→Wsuch that T(v) = wand T(v0) = 0 for v0∈T. But

for such a T, we have Ev(T) = T(v) = w. Since w∈Wwas arbitrary, this means that Ev

is onto.

5. Let Fbe a field and Van F-vector space. The vector space L(V, F ) is called the dual

vector space of Vand also written V∗. Prove that dimF(V) = dimF(V∗) if Vis finite

dimensional. (4 pts)

Choose an ordered basis Aof Vand an ordered basis (i.e., a non-zero element) Bof F. Then

the matrix representation [ ]B

A:V∗→M(1 ×n, F ) is an isomorphism. The latter space is

simply the space of row vectors of length n, which has dimension n. Therefore V∗also has

dimension n. (Remark: If Vis not finite-dimensional, then there is no isomorphism between

Vand V∗- the dual vector space is ”larger” than the vector space.)

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Math 115A Homework 3 Due November 7th, 2008

Let Id : R^3 → R^3 be the identity linear transformation. For the ordered bases A, B, compute the matrix representation [Id]BA. (To be clear, the bases are ordered by reading from left to right.) (2 pts each)

a) A = { 2 e 2 , e 3 − e 1 , e 1 } and B = {e 1 , 2 e 2 , e 3 − e 1 }.

b) A = {e 1 , e 1 + e 2 , e 1 + e 2 + e 3 } and B = { 2 e 1 − 3 e 2 , e 3 + e 2 , 4 e 2 }.

Prove: If F is a field, V is an F -vector space of dimension n, and A is any ordered basis of V , then the matrix representation [Id]AA of the identity linear transformation is the n × n-identity matrix In. (4 pts)

This is essentially obvious. Let A = {v 1 ,... , vn}. The i-th column of the matrix [Id]AA is the column vector [Id(vi)]A = ei. That is precisely the i-th column of the identity matrix.

Prove: Let F be a field and let V be an F -vector space of dimension n. Suppose T : V → V is a linear transformation of rank n. Then there are ordered bases A and B of V such that [T ]BA = In is the n × n-identity matrix. (4 pts)

Choose any ordered basis A = {v 1 ,... , vn}. The set {T (v 1 ),... , T (vn)} spans the range of T , which is V ; since V has dimension n, B = {T (v 1 ),... , T (vn)} is an ordered basis of V. The i-th column of [T ]BA is [T (vi)]B = ei, hence [T ]BA = In.

Let F be a field and let V and W be F -vector spaces. Suppose v ∈ V. Prove that the map Ev : L(V, W ) → W defined by Ev(T ) = T (v) is a linear transformation. Moreover, show that Ev is onto provided v 6 = 0. (You may assume that V has finite dimension.) (4 pts)

Let S and T be linear transformations from V to W and let λ ∈ F be a scalar. Then

Ev(S + λT ) = (S + λT )(v) = S(v) + λT (v) = Ev(S) + λEv(T )

where the first equality is by definition of the addition and scalar multiplication on L(V, W ). That is, Ev is linear. Now suppose v 6 = 0. Suppose w ∈ W. Let S ⊆ V be any basis of V. By the replacement theorem, we can find a subset T ⊆ S such that B := T ∪ {v} is a basis of V. Now there is a unique linear transformation T : V → W such that T (v) = w and T (v′) = 0 for v′^ ∈ T. But for such a T , we have Ev(T ) = T (v) = w. Since w ∈ W was arbitrary, this means that Ev is onto.

Let F be a field and V an F -vector space. The vector space L(V, F ) is called the dual vector space of V and also written V ∗. Prove that dimF (V ) = dimF (V ∗) if V is finite dimensional. (4 pts)

Choose an ordered basis A of V and an ordered basis (i.e., a non-zero element) B of F. Then the matrix representation [ ]BA : V ∗^ → M (1 × n, F ) is an isomorphism. The latter space is simply the space of row vectors of length n, which has dimension n. Therefore V ∗^ also has dimension n. (Remark: If V is not finite-dimensional, then there is no isomorphism between V and V ∗^ - the dual vector space is ”larger” than the vector space.)

Math 115A Homework 3: Linear Transformations and Matrix Representations, Assignments of Linear Algebra

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