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Chapter 4
Linear Transformations

4.1 Definitions and Basic Properties

Let V be a vector space over F with dim(V) = n. Also, let B be an ordered basis of V. Then, in the last section of the previous chapter, it was shown that for each x ∈ V, the coordinate vector [x]B is a column vector of size n and has entries from F. So, in some sense, each element of V looks like elements of Fn. In this chapter, we concretize this idea. We also show that matrices give rise to functions between two finite dimensional vector spaces. To do so, we start with the definition of functions over vector spaces that commute with the operations of vector addition and scalar multiplication.

Definition 4.1.1. [Linear Transformation, Linear Operator] Let V and W be vector spaces over F. A function (map) T : V → W is called a linear transformation if for all α ∈ F and u,v ∈ V the function T satisfies

T(α ⋅u) = α ⊙ T(u) and T (u + v ) = T (u )⊕ T (v ),
where +,⋅ are binary operations in V and ⊕,⊙ are the binary operations in W. By L(V, W), we denote the set of all linear transformations from V to W. In particular, if W = V then the linear transformation T is called a linear operator and the corresponding set of linear operators is denoted by L(V).

Definition 4.1.2. [Equality of Linear Transformation] Let S,T ∈L(V, W). Then, S and T are said to be equal if S(x) = T(x), for all x ∈ V. PICT PICT DRAFT

We now give examples of linear transformations.

Example 4.1.3.

1.
Let V be a vector space. Then, the maps Id,0 ∈L(V), where
(a)
Id(v) = v, for all v ∈ V, is commonly called the identity operator.
(b)
0(v) = 0, for all v ∈ V, is commonly called the zero operator.
2.
Let V and W be two vector spaces over F. Then, 0 ∈L(V, W), where 0(v) = 0, for all v ∈ V, is commonly called the zero transformation.
3.
The map T(x) = x, for all x ∈ ℝ, is an element of L(ℝ) as T(ax) = ax = aT(x) and T(x + y) = x + y = T(x) + T(y).
4.
The map T(x) = (x,3x)T , for all x ∈ ℝ, is an element of L(ℝ, ℝ2) as T(λx) = (λx,3λx)T = λ(x,3x)T = λT(x) and T(x + y) = (x + y,3(x + y)T = (x,3x)T + (y,3y)T = T(x) + T(y).
5.
Let V, W and ℤ be vector spaces over F. Then, for any T ∈L(V, W) and S ∈L(W, ℤ), the map S ∘T ∈L(V, ℤ), where (S ∘T)(v) = S(T(v)), for all v ∈ V, is called the composition of maps. Observe that for each v ∈ V,
(S ∘T )(αv + βu) = S (T(αv + βu )) = S (αT (v) + βT(u )) = αS (T(v )) + βS(T (u)) = α (S ∘ T)(v) + β(S ∘T )(u )
and hence S ∘ T, in short ST, is an element of L(V, ℤ). PICT PICT DRAFT
6.
Fix a ∈ ℝn and define T(x) = aT x, for all x ∈ ℝn. Then T ∈L(ℝn, ℝ). For example,
(a)
if a = (1,…,1)T then T(x) = ∑ i=1nxi, for all x ∈ ℝn.
(b)
if a = ei, for a fixed i,1 ≤ i ≤ n, then Ti(x) = xi, for all x ∈ ℝn.
7.
Define T : ℝ2 → ℝ3 by T( T) (x,y) = (x + y,2x - y,x + 3y)T . Then T ∈L(ℝ2, ℝ3) with T(e1) = (1,2,1)T and T(e2) = (1,-1,3)T .
8.
Let A ∈ Mm×n(ℂ). Define TA(x) = Ax, for every x ∈ ℂn. Then, TA ∈L(ℂn, ℂm). Thus, for each A ∈ Mm,n(ℂ), there exists a linear transformation TA ∈L(ℂn, ℂm).
9.
Define T : ℝn+1 → ℝ[x;n] by T( T) (a1,...,an+1) = a1 + a2x + ⋅⋅⋅ + an+1xn, for each (a1,…,an+1) ∈ ℝn+1. Then T is a linear transformation.
10.
Fix A ∈ Mn(ℂ). Now, define TA : Mn(ℂ) → Mn(ℂ) and SA : Mn(ℂ) → ℂ by TA(B) = AB and SA(B) = Tr(AB), for every B ∈ Mn(ℂ). Then, TA and SA are both linear transformations. What can you say about the maps f1(B) = A*B,f2(B) = BA,f3(B) = tr(A*B) and f4(B) = tr(BA), for every B ∈ Mn(ℂ)?
11.
Verify that the map T : ℝ[x;n] → ℝ[x;n + 1] defined by T(f(x)) = xf(x), for all f(x) ∈ ℝ[x;n] is a linear transformation.
12.
The maps T,S : ℝ[x] → ℝ[x] defined by T(f(x)) = -d dxf(x) and S(f(x)) = ∫ 0xf(t)dt, for all f(x) ∈ ℝ[x] are linear transformations. Is it true that TS = Id? What about ST?
13.
Recall the vector space ℝℕ in Example 3.1.4.8. Now, define maps T,S : ℝℕ → ℝℕ by T({a1,a2,…}) = {0,a1,a2,…} and S({a1,a2,…}) = {a2,a3,…}. Then, T and S, are commonly called the shift operators, are linear operators with exactly one of ST or TS as the Id map.
14.
Recall the vector space C(ℝ, ℝ) (see Example 3.1.4.10). We now define a map T : C(ℝ, ℝ) →C(ℝ, ℝ) by T(f(x)) = ∫ 0xf(t)dt. For example, (T(sin(x)) = ∫ 0x sin(t)dt = 1 - cos(x), for all x ∈ ℝ. Then, verify that T is a linear transformation.
PICT PICT DRAFT

Remark 4.1.4. Let A ∈ Mn(ℂ) and define TA : ℂn → ℂn by TA(x) = Ax, for every x ∈ ℂn. Then, verify that TAk(x) = (◟TA-∘-TA◝∘◜⋅⋅⋅∘TA-)◞k times(x) = Akx, for any positive integer k.

Also, for any two linear transformations S ∈ L(V, W) and T ∈ L(W, ℤ), we will interchangeably use T ∘ S and TS, for the corresponding linear transformation in L(V, ℤ).

We now prove that any linear transformation sends the zero vector to a zero vector.

Proposition 4.1.5. Let T ∈L(V, W). Suppose that 0

V isthezerovectorinVand0˙WisthezerovectorofW.ThenT(0˙V) = 0˙W.

Proof. Since 0

V = 0˙V + 0˙V,weget T(0˙V) = T(0˙V + 0˙V)= T(0˙V) + T(0˙V).AsT(0˙V) ∈ W,

0W + T (0V) = T(0V) = T(0V )+ T(0V ).
Hence, T(0

V) = 0˙W. _

From now on 0 will be used as the zero vector of the domain and codomain. We now consider a few more examples.

Example 4.1.6. PICT PICT DRAFT

1.
Does there exist a linear transformation T : V → W such that T(v)≠0, for all v ∈ V?
Solution: No, as T(0) = 0 (see Proposition 4.1.5).
2.
Does there exist a linear transformation T : ℝ → ℝ such that T(x) = x2, for all x ∈ ℝ?
Solution: No, as T(ax) = (ax)2 = a2x2 = a2T(x)≠aT(x), unless a = 0,1.
3.
Does there exist a linear transformation T : ℝ → ℝ such that T(x) = √x--, for all x ∈ ℝ?
Solution: No, as T(ax) = √ --- ax = √ -- a√ -- x≠a√ -- x = aT(x), unless a = 0,1.
4.
Does there exist a linear transformation T : ℝ → ℝ such that T(x) = sin(x), for all x ∈ ℝ?
Solution: No, as T(ax)≠aT(x).
5.
Does there exist a linear transformation T : ℝ → ℝ such that T(5) = 10 and T(10) = 5?
Solution: No, as T(10) = T(5 + 5) = T(5) + t(5) = 10 + 10 = 20≠5.
6.
Does there exist a linear transformation T : ℝ → ℝ such that T(5) = π and T(e) = π?
Solution: No, as 5T(1) = T(5) = π implies that T(1) = π- 5. So, T(e) = eT(1) = eπ- 5.
7.
Does there exist a linear transformation T : ℝ2 → ℝ2 such that T((x,y)T ) = (x + y,2)T ?
Solution: No, as T(0)≠0.
8.
Does there exist a linear transformation T : ℝ2 → ℝ2 such that T((x,y)T ) = (x+y,xy)T ?
Solution: No, as T((2,2)T ) = (4,4)T ≠2(2,1)T = 2T((1,1)T ).
9.
Define a map T : ℂ → ℂ by T(z) = z, the complex conjugate of z. Is T a linear operator over the real vector space ℂ?
Solution: Yes, as for any α ∈ ℝ, T(αz) = αz = αz = αT(z).

The next result states that a linear transformation is known if we know its image on a basis of the domain space.

PICT PICT DRAFT Lemma 4.1.7. Let V and W be two vector spaces over F and let T ∈ L(V, W). Then T is known, if the image of T on basis vectors of V are known. In particular, if V is finite dimensional and B = (v1,…,vn) is an ordered basis of V over F then, T(v) = [ ] T (v1) ⋅⋅⋅ T(vn)[v]B.

Proof. Let B be a basis of V over F. Then, for each v ∈ V, there exist vectors u1,…,uk in B and scalars c1,…,ck ∈ F such that v = ∑ i=1kciui. Thus, by definition T(v) = ∑ i=1kciT(ui). Or equivalently, whenever

⌊ ⌋ ⌊ ⌋ c1 [ ] c1 v = [u1, ...,uk ]|| ..|| then, T(v) = T(u ) ⋅⋅⋅ T(u ) || ..|| . ⌈ .⌉ 1 k ⌈ .⌉ ck ck
(4.1.1)

Thus, the image of T on v just depends on where the basis vectors are mapped. In particular, if [v]B = ⌊ ⌋ c1 || .|| ⌈ ..⌉ cn then, T(v) = [ ] T (u1) ⋅⋅⋅ T(uk )[v]B. Hence, the required result follows. _

As another application of Lemma 4.1.7, we have the following result. The proof is left for the reader.

Corollary 4.1.8. Let V and W be vector spaces over F and let T : V → W be a linear transformation. If B is a basis of V then, Rng(T) = LS(T(x)|x ∈B).

Recall that by Example 4.1.3.6, for each a ∈ Fn, the map T(x) = aT x, for each x ∈ Fn, is a linear transformation. We now show that these are the only ones.

PICT PICT DRAFT Corollary 4.1.9. [Reisz Representation Theorem] Let T ∈ L(ℝn, ℝ). Then, there exists a ∈ ℝn such that T(x) = aT x.

Proof. By Lemma 4.1.7, T is known if we know the image of T on {e1,…,en}, the standard basis of ℝn. As T is given, for 1 ≤ i ≤ n, T(ei) = ai, for some ai ∈ ℝ. So, consider the vector a = [a1,…,an]T . Then, for x = [x1,…,xn]T ∈ ℝn, we see that

( ) ∑n ∑n ∑n T T(x) = T xiei = xiT(ei) = xiai = a x. i=1 i=1 i=1
Thus, the required result follows. _

Before proceeding further, we define two spaces related with a linear transformation.

Definition 4.1.10. [Range and Kernel of a Linear Transformation] Let V and W be vector spaces over F and let T : V → W be a linear transformation. Then,

1.
the set {T(v)|v ∈ V} is called the range space of T, denoted Rng(T).
2.
the set {v ∈ V|T(v) = 0} is called the kernel of T, denoted Ker(T). In certain books, it is also called the null space of T.

Example 4.1.11. Determine Rng(T) and Ker(T) of the following linear transformations.

1.
T ∈L(ℝ3, ℝ4), where T((x,y,z)T ) = (x - y + z,y - z,x,2x - 5y + 5z)T .
Solution: Consider the standard basis {e1,e2,e3} of ℝ3. Then
PICT PICT DRAFT
T T T Rng (T) = LS(T (e1),T (e2),T (e3)) = LS ((1,0,1,2 ) ,(- 1,1,0,- 5 ) ,(1,- 1,0,5) ) = LS((1,0,1,2)T,(1,- 1,0,5)T) = {λ(1,0,1,2)T + β(1,- 1,0,5)T|λ,β ∈ ℝ } = {(λ + β,- β,λ,2λ + 5β ) : λ,β ∈ ℝ } = {(x, y,z,w)T ∈ ℝ4|x+ y - z = 0,5y - 2z + w = 0}
and
T 3 T Ker (T ) = {(x,y,z) ∈ ℝ : T((x,y,z) ) = 0} = {(x,y,z)T ∈ ℝ3 : (x- y + z,y - z,x,2x - 5y + 5z)T = 0 } = {(x,y,z)T ∈ ℝ3 : x - y + z = 0,y - z = 0,x = 0,2x- 5y + 5z = 0} T 3 = {(x,y,z) ∈ ℝ : y - z = 0,x = 0} = {(0,z,z)T ∈ ℝ3 : z ∈ ℝ } = LS ((0,1,1)T)
2.
Let B ∈ M2(ℝ). Now, define a map T : M2(ℝ) → M2(ℝ) by T(A) = BA - AB, for all A ∈ M2(ℝ). Determine Rng(T) and Ker(T).
Solution: Note that A ∈ Ker(T) if and only if A commutes with B. In particular, {I,B,B2,…}⊆ Ker(T). For example, if B is a scalar matrix then, Ker(T) = M2(ℝ).

For computing, Rng(T), recall that {Eij|1 ≤ i,j ≤ 2} is a basis of M2(ℝ). So, for example,

(a)
for B = cI2, verify that Rng(T) = {0}.
(b)
for B = [ ] 1 2 2 4, verify that PICT PICT DRAFT
( [ ] [ ] [ ]) 0 2 2 3 - 2 0 Rng (T ) = LS (T(Eij),1 ≤ i,j ≤ 2) = LS - 2 0 , 0 - 2 , - 3 2 .
(c)
for B = [ ] 1 2 2 3, verify that
( [ ] [ ] [ ]) Rng (T ) = LS (T(Eij),1 ≤ i,j ≤ 2) = LS 0 2 , 2 2 , - 2 0 . - 2 0 0 - 2 - 2 2

Exercise 4.1.12.

1.
Let V and W be two vector spaces over F. If {v1,…,vn} is a basis of V and w1,…,wn ∈ W then prove that there exists a unique T ∈L(V, W) such that T(vi) = wi, for i = 1,…,n.
2.
Let V and W be two vector spaces over F and let T ∈L(V, W). Then
(a)
Rng(T) is a subspace of W.
(b)
Ker(T) is a subspace of V.

Furthermore, if V is finite dimensional then PICT PICT DRAFT

(a)
dim(Ker(T)) ≤ dim(V).
(b)
dim(Rng(T)) is finite and whenever dim(W) is finite dim(Rng(T)) ≤ dim(W).
3.
Describe Ker(D) and Rng(D), where D ∈L(ℝ[x;n]) and is defined by D(f(x)) = f′(x). Note that Rng(D) ⊆ ℝ[x;n - 1].
4.
Define T ∈L(ℝ[x]) by T(f(x)) = xf(x), for all f(x) ∈L(ℝ[x]). What can you say about Ker(T) and Rng(T)?
5.
Determine the dimension of Ker(T) and Rng(T), for the linear transformation T, given in Example 4.1.11.2. What can you say if B is skew-symmetric?_________________________
6.
Define T ∈L(ℝ3) by T(e1) = e1 + e3, T(e2) = e2 + e3 and T(e3) = -e3.
(a)
Then, determine T((x,y,z)T ), for x,y,z ∈ ℝ.
(b)
Then, determine Null(T) and Rng(T).
(c)
Then, is it true that T2 = Id?
7.
Find T ∈L(ℝ3) for which Rng(T) = LS((1,2,0)T ,(0,1,1)T ,(1,3,1)T ).

Example 4.1.13. In each of the examples given below, state whether a linear transformation exists or not. If yes, give at least one linear transformation. If not, then give the condition due to which a linear transformation doesn’t exist.

1.
T : ℝ2 → ℝ2 such that T((1,1)T ) = (1,2)T and T((1,-1)T ) = (5,10)T ?
Solution: Yes, as the set {(1,1),(1,-1)} is a basis of ℝ2, the matrix [ ] 1 1 1 - 1 is invertible. Also, T([ ][ ]) 1 1 a 1 - 1 b = T( [ ] [ ]) 1 1 a + b 1 - 1 = a[ ] 1 2 + b[ ] 5 10 = [ ] 1 5 2 10[ ] a b. So,
PICT PICT DRAFT
( [ ]) ( [ ]( [ ] [ ]) ) [ ]( [ ] [ ]) x 1 1 1 1 - 1 x 1 5 1 1 -1 x T = T ( ( ) ) = ( ) y 1 - 1 1 - 1 y 2 10 1 - 1 y [ ][x+y-] [ ] = 1 5 2 = 3x- 2y . 2 10 x-2y- 6x- 4y
2.
T : ℝ2 → ℝ2 such that T((1,1)T ) = (1,2)T and T((5,5)T ) = (5,10)T ?
Solution: Yes, as (5,10)T = T((5,5)T ) = 5T((1,1)T ) = 5(1,2)T = (5,10)T .

To construct one such linear transformation, let {(1,1)T ,u} be a basis of ℝ2 and define T(u) = v = (v1,v2)T , for some v ∈ ℝ2. For example, if u = (1,0)T then

( [ ]) ( [ ]( [ ]- 1[ ]) ) [ ]( [ ]- 1[ ]) [ ] x 1 1 1 1 x 1 v1 1 1 x 1 T y = T( 1 0 ( 1 0 y ) ) = 2 v ( 1 0 y ) = y 2 + (x- y)v. 2
3.
T : ℝ2 → ℝ2 such that T((1,1)T ) = (1,2)T and T((5,5)T ) = (5,11)T ?
Solution: No, as (5,11)T = T((5,5)T ) = 5T((1,1)T )5(1,2)T = (5,10)T , a contradiction.
4.
T : ℝ2 → ℝ2 such that Rng(T) = {T(x)|x ∈ ℝ2} = LS{(1,π)T }?
Solution: Yes. Define T(e1) = (1,π)T and T(e2) = 0 or T(e1) = (1,π)T and T(e2) = a(1,π)T , for some a ∈ ℝ.
5.
T : ℝ2 → ℝ2 such that Rng(T) = {T(x)|x ∈ ℝ2} = ℝ2?
Solution: Yes. Define T(e1) = (1,π)T and T(e2) = (π,e)T . Or, let {u,v} be a basis of ℝ2 and define T(e1) = u and T(e2) = v. PICT PICT DRAFT
6.
T : ℝ2 → ℝ2 such that Rng(T) = {T(x)|x ∈ ℝ2} = {0}?
Solution: Yes. Define T(e1) = 0 and T(e2) = 0.
7.
T : ℝ2 → ℝ2 such that Ker(T) = {x ∈ ℝ2|T(x) = 0} = LS{(1,π)T }?
Solution: Yes. Let a basis of ℝ2 = {(1,π)T ,(1,0)T } and define T((1,π)T ) = 0 and T((1,0)T ) = u≠0.

Exercise 4.1.14.

1.
Which result gives the statement “Prove that a map T : ℝ → ℝ is a linear transformation if and only if there exists a unique c ∈ ℝ such that T(x) = cx, for every x ∈ ℝ”?
2.
Use matrices to give examples of linear operators T,S : ℝ3 → ℝ3 that satisfy:
(a)
T≠0,T ∘ T = T2≠0,T ∘ T ∘ T = T3 = 0.
(b)
T≠0,S≠0,S ∘ T = ST≠0,T ∘ S = TS = 0.
(c)
S ∘ S = S2 = T2 = T ∘ T,S≠T.
(d)
T ∘ T = T2 = Id,T≠ Id.

From now on, we will just write

3.
Let T : ℝn → ℝn be a linear operator with T≠0 and T2 = 0. Prove that there exists a vector x ∈ ℝn such that the set {x,T(x)} is linearly independent.
4.
Fix a positive integer p and let T : ℝn → ℝn be a linear operator with Tk≠0 for 1 ≤ k ≤ p and Tp+1 = 0. Then prove that there exists a vector x ∈ ℝn such that the set {x,T(x),…,Tp(x)} is linearly independent.
5.
Let T : ℝn → ℝm be a linear transformation with T(x0) = y0, for some x0 ∈ ℝn and y0 ∈ ℝm. PICT PICT DRAFT Define T-1(y0) = {x ∈ ℝn : T(x) = y0}. Then prove that for every x ∈ T-1(y0) there exists z ∈ T-1(0) such that x = x0 + z. Also, prove that T-1(y0) is a subspace of ℝn if and only if 0 = y0.__________________________________________________________________________
6.
Prove that there exists infinitely many linear transformations T : ℝ3 → ℝ2 such that T((1,-1,1)T ) = (1,2)T and T((-1,1,2)T ) = (1,0)T ?
7.
Let V be a vector space and let a ∈ V. Then the map Ta : V → V defined by Ta(x) = x + a, for all x ∈ V is called the translation map. Prove that Ta ∈L(V) if and only if a = 0.
8.
Are the maps T : V → W given below, linear transformations? In case, T is a linear transformation, determine Rng(T) and Ker(T).
(a)
Let V = ℝ2 and W = ℝ3 with T((x,y)T ) = (x + y + 1,2x - y,x + 3y)T .
(b)
Let V = W = ℝ2 with T((x,y)T ) = (x - y,x2 - y2)T .
(c)
Let V = W = ℝ2 with T((x,y)T ) = (x - y,|x|)T .
(d)
Let V = ℝ2 and W = ℝ4 with T((x,y)T ) = (x + y,x - y,2x + y,3x - 4y)T .
(e)
Let V = W = ℝ4 with T((x,y,z,w)T ) = (z,x,w,y)T .
9.
Which of the following maps T : M2(ℝ) → M2(ℝ) are linear operators? In case, T is a linear operator, determine Rng(T) and Ker(T).
(a)
T(A) = AT .
(b)
T(A) = I + A.
(c)
T(A) = A2.
(d)
T(A) = BAB-1, for some fixed B ∈ M2(ℝ).
10.
Does there exist a linear transformation T : ℝ3 → ℝ2 such that
(a)
T((1,0,1)T ) = (1,2)T , T((0,1,1)T ) = (1,0)T and T((1,1,1)T ) = (2,3)T ? PICT PICT DRAFT
(b)
T((1,0,1)T ) = (1,2)T , T((0,1,1)T ) = (1,0)T and T((1,1,2)T ) = (2,3)T ?
11.
Let T : ℝ3 → ℝ3 be defined by T((x,y,z)T ) = (2x + 3y + 4z,x + y + z,x + y + 3z)T . Find the value of k for which there exists a vector x ∈ ℝ3 such that T(x) = (9,3,k)T .
12.
Let T : ℝ3 → ℝ3 be defined by T((x,y,z)T ) = (2x- 2y + 2z,-2x + 5y + 2z,8x + y + 4z)T . Find x ∈ ℝ3 such that T(x) = (1,1,-1)T .
13.
Let T : ℝ3 → ℝ3 be defined by T((x,y,z)T ) = (2x + y + 3z,4x - y + 3z,3x - 2y + 5z)T . Determine x,y,z ∈ ℝ3 \{0} such that T(x) = 6x, T(y) = 2y and T(z) = -2z. Is the set {x,y,z} linearly independent?
14.
Let T : ℝ3 → ℝ3 be defined by T((x,y,z)T ) = (2x + 3y + 4z,-y,-3y + 4z)T . Determine x,y,z ∈ ℝ3 \{0} such that T(x) = 2x, T(y) = 4y and T(z) = -z. Is the set {x,y,z} linearly independent?
15.
Let n ∈ ℕ. Does there exist a linear transformation T : ℝ3 → ℝn such that T((1,1,-2)T ) = x, T((-1,2,3)T ) = y and T((1,10,1)T ) = z
(a)
with z = x + y?
(b)
with z = cx + dy, for some c,d ∈ ℝ?
16.
For each matrix A given below, define T ∈L(ℝ2) by T(x) = Ax. What do these linear operators signify geometrically?
(a)
A ∈ { [ ] [ ] [ ] [ ] [ ]} 1 √3-- - 1 1 1 - 1 1 1 - √3- 0 - 1 cos(2π) - sin (2π) -- √ --, √-- ,--√ -- , , (23π) (2π3) 2 1 3 2 1 1 2 3 1 1 0 sin 3 cos 3.
(b)
A ∈{[- 1 0] [1 0 ] 1 [1 1] 1[1 2] [0 0] [1 0]} , ,-- ,-- , , 0 1 0 - 1 2 1 - 1 5 2 4 0 1 0 0.
(c)
A ∈{ [√ -- ] [ ] [ √--] [ (2π) (2π) ] } 1- 3 1√ --,√1-- 1 1 , 1-√1-- 3 , cos( 3) sin (3 ) 2 1 - 3 2 1 - 1 2 3 - 1 sin 2π3 - cos 2π3.
17.
Find all functions f : ℝ2 → ℝ2 that fixes the line y = x and sends (x1,y1) for x1≠y1 to its mirror image along the line y = x. Or equivalently, f satisfies
(a)
f(x,x) = (x,x) and PICT PICT DRAFT
(b)
f(x,y) = (y,x) for all (x,y) ∈ ℝ2.
18.
Consider the space ℂ3 over ℂ. If f ∈L(ℂ3) with f(x) = x,f(y) = (1 + i)y and f(z) = (2 + 3i)z, for x,y,z ∈ ℂ3 \{0} then prove that {x,y,z} form a basis of ℂ3.

4.2 Rank-Nullity Theorem

The readers are advised to see Exercise 3.3.13.9 and Theorem 3.5.9 for clarity and similarity with the results in this section. We start with the following result.

Theorem 4.2.1. Let V and W be two vector spaces over F and let T ∈L(V, W).

1.
If S ⊆ V is linearly dependent then T(S) = {T(v)|v ∈ V} is linearly dependent.
2.
Suppose S ⊆ V such that T(S) is linearly independent then S is linearly independent.

Proof. As S is linearly dependent, there exist k ∈ ℕ and vi ∈ S, for 1 ≤ i ≤ k, such that the system ∑ i=1kxivi = 0, in the variable xi’s, has a non-trivial solution, say xi = ai ∈ F,1 ≤ i ≤ k. Thus, ∑ i=1kaivi = 0. Now, consider the system ∑ i=1kyiT(vi) = 0, in the variable yi’s. Then,

( ) ∑k ∑k ∑ k aiT (vi) = T (aivi) = T aivi = T(0) = 0. i=1 i=1 i=1
Thus, ai’s give a non-trivial solution of ∑ i=1kyiT(vi) = 0 and hence the required result follows.

The second part is left as an exercise for the reader. _ PICT PICT DRAFT

Definition 4.2.2. [Rank and Nullity] Let V and W be two vector spaces over F. If T ∈ L(V, W) and dim(V) is finite then we define Rank(T) = dim(Rng(T)) and Nullity(T) = dim(Ker(T)).

We now prove the rank-nullity Theorem. The proof of this result is similar to the proof of Theorem 3.5.9. We give it again for the sake of completeness.

Theorem 4.2.3 (Rank-Nullity Theorem). Let V and W be two vector spaces over F. If dim(V) is finite and T ∈L(V, W) then,

Rank (T) + Nullity (T) = dim(Rng (T ))+ dim (Ker (T )) = dim (V).

Proof. By Exercise 4.1.12.2.2a, dim(Ker(T)) ≤ dim(V). Let B be a basis of Ker(T). We extend it to form a basis C of V. As, T(v) = 0, for all v ∈B, using Corollary 4.1.8, we get

Rng (T) = LS ({T(v)|v ∈ C}) = LS ({T (v )|v ∈ C \B }).
We claim that {T(v)|v ∈C\B$ is linearly independent subset of W.

Let, if possible, the claim be false. Then, there exists v1,…,vk ∈C\B and a = [a1,…,ak]T such that a≠0 and ∑ i=1kaiT(vi) = 0. Thus, we see that PICT PICT DRAFT

( k ) k T ∑ a v = ∑ a T (v) = 0. i=1 i i i=1 i i
That is, ∑ i=1kaivi ∈ Ker(T). Hence, there exists b1,…,bℓ ∈ F and u1,…,uℓ ∈B such that ∑ i=1kaivi = ∑ j=1kbjuj. Or equivalently, the system ∑ i=1kxivi + ∑ j=1kyjuj = 0, in the variables xi’s and yj’s, has a non-trivial solution [a1,…,ak,-b1,…,-bℓ]T (non-trivial as a≠0). Hence, S = {v1,…,vk,u1,…,uℓ} is linearly dependent subset in V. A contradiction to S ⊆C. That is,
dim(Rng (T ))+ dim(Ker (T )) = |C \ B|+ |B| = |C | = dim(V ).
Thus, we have proved the required result. _

As an immediate corollary, we have the following result. The proof is left for the reader.

Corollary 4.2.4. Let V and W be vector spaces over F and let T ∈L(V, W). If dim(V) = dim(W) then, the following statements are equivalent.

1.
T is one-one.
2.
Ker(T) = {0}.
3.
T is onto.
4.
dim(Rng(T)) = dim(V).
PICT PICT DRAFT

Corollary 4.2.5. Let V be a vector space over F with dim(V) = n. If S,T ∈L(V). Then

1.
Nullity(T) + Nullity(S) ≥ Nullity(ST) ≥ max{Nullity(T),Nullity(S)}.
2.
min{Rank(S),Rank(T)}≥Rank(ST) ≥ n -Rank(S) -Rank(T).

Proof. The prove of Part 2 is omitted as it directly follows from Part 1 and Theorem 4.2.3.
Part 1: We first prove the second inequality. Suppose v ∈ Ker(T). Then

(ST )(v) = S(T (v )) = S (0 ) = 0
implies Ker(T) ⊆ Ker(ST). Thus, Nullity(T) ≤ Nullity(ST).

By Theorem 4.2.3, Nullity(S) ≤ Nullity(ST) is equivalent to Rng(ST) ⊆ Rng(S). And this holds as Rng(T) ⊆ V implies Rng(ST) = S(Rng(T)) ⊆ S(V) = Rng(S).

To prove the first inequality, let {v1,…,vk} be a basis of Ker(T). Then {v1,…,vk}⊆ Ker(ST). So, let us extend it to get a basis {v1,…,vk,u1,…,uℓ} of Ker(ST).

Claim: {T(u1),T(u2),…,T(uℓ)} is a linearly independent subset of Ker(S).

Clearly, {T(u1),…,T(uℓ)}⊆ Ker(S). Now, consider the system c1T(u1) + ⋅⋅⋅ + cℓT(uℓ) = 0 in the variables c1,…,cℓ. As T ∈L(V), we get T( ) ∑ℓ cu i=1 i i = 0. Thus, ∑ i=1ℓciui ∈ Ker(T). Hence, ∑ i=1ℓciui is a unique linear combination of v1,…,vk, a basis of Ker(T). Therefore,

c1u1 + ⋅⋅⋅+ cℓuℓ = α1v1 + ⋅⋅⋅+ αkvk
(4.2.1)

for some scalars α1,…,αk. But by assumption, {v1,…,vk,u1,…,uℓ} is a basis of Ker(ST) and hence linearly independent. Therefore, the only solution of Equation (4.2.1) is given by ci = 0, for 1 ≤ i ≤ ℓ and αj = 0, for 1 ≤ j ≤ k. Thus, we have proved the claim. Hence, Nullity(S) ≥ ℓ and Nullity(ST) = k + ℓ ≤ Nullity(T) + Nullity(S). _

Exercise 4.2.6.

1.
Let A ∈ Mn(ℝ) with A2 = A. Define T ∈L(ℝn, ℝn) by T(v) = Av, for all v ∈ ℝn.
(a)
Then, prove that T2 = T. Equivalently, T(Id - T) = 0. That is, prove that (T(Id - T))(x) = 0, for all x ∈ ℝn.
(b)
Then, prove that Null(T) ∩ Rng(T) = {0}.
(c)
Then, prove that ℝn = Rng(T) + Null(T).
____________________________________
2.
Define T ∈L(ℝ3, ℝ2) by T( ⌊x⌋ ) | | | | ( ⌈y⌉ ) z = [ ] x- y + z x + 2z. Find a basis and the dimension of Rng(T) and Ker(T).
3.
Let zi ∈ ℂ, for 1 ≤ i ≤ k. Define T ∈L(ℂ[x;n], ℂk) by T(P(z)) = (P(z1),…,P(zk)). If zi’s are distinct then for each k ≥ 1, determine Rank(T).

4.2.1 Algebra of Linear Transformations

We start with the following definition.

PICT PICT DRAFT Definition 4.2.7. [Sum and Scalar Multiplication of Linear Transformations] Let V, W be vector spaces over F and let S,T ∈L(V, W). Then, we define the point-wise

1.
sum of S and T, denoted S + T, by (S + T)(v) = S(v) + T(v), for all v ∈ V.
2.
scalar multiplication, denoted cT for c ∈ F, by (cT)(v) = c(T(v)), for all v ∈ V.

Theorem 4.2.8. Let V and W be vector spaces over F. Then L(V, W) is a vector space over F. Furthermore, if dim V = n and dim W = m, then dimL(V, W) = mn.

Proof. It can be easily verified that for S,T ∈L(V, W), if we define (S + αT)(v) = S(v) + αT(v) (point-wise addition and scalar multiplication) then L(V, W) is indeed a vector space over F. We now prove the other part. So, let us assume that B = {v1,…,vn} and C = {w1,…,wm} are bases of V and W, respectively. For 1 ≤ i ≤ n,1 ≤ j ≤ m, we define the functions fij on the basis vectors of V by

{ wj, if k = i fij(vk) = 0, if k ⁄= i.
For other vectors of V, we extend the definition by linearity. That is, if v = ∑ s=1nαsvs then,
( n ) n ∑ ∑ fij(v) = fij αsvs = αsfij(vs) = αifij(vi) = αiwj. s=1 s=1
(4.2.2)

Thus, fij ∈L(V, W).

Claim: {fij|1 ≤ i ≤ n,1 ≤ j ≤ m} is a basis of L(V, W).

So, consider the linear system ∑ i=1n ∑ j=1mcijfij = 0, in the variables cij’s, for 1 ≤ i ≤ n,1 ≤ j ≤ m. Using the point-wise addition and scalar multiplication, we get

( n m ) n m m (∑ ∑ ) ∑ ∑ ∑ 0 = 0(vk) = cijfij (vk ) = cijfij(vk ) = ckjwj. i=1 j=1 i=1 j=1 j=1
But, the set {w1,…,wm} is linearly independent and hence the only solution equals ckj = 0, for 1 ≤ j ≤ m. Now, as we vary vk from v1 to vn, we see that cij = 0, for 1 ≤ j ≤ m and 1 ≤ i ≤ n. Thus, we have proved the linear independence.

Now, let us prove that LS({fij|1 ≤ i ≤ n,1 ≤ j ≤ m }) = L(V, W). So, let f ∈L(V, W). Then, for 1 ≤ s ≤ n, f(vs) ∈ W and hence there exists βst’s such that f(vs) = ∑ t=1mβstwt. So, if v = ∑ s=1nαsvs ∈ V then, using Equation (4.2.2), we get

( n ) n n ( m ) n m ∑ ∑ ∑ ∑ ∑ ∑ f(v) = f αsvs = αsf (vs ) = αs βstwt = βst(αswt) s=1 s=1( s=1) t=1 s=1 t=1 ∑n ∑m ∑n ∑m = βstfst(v) = βstfst (v). s=1 t=1 s=1 t=1
Since the above is true for every v ∈ V, LS({fij|1 ≤ i ≤ n,1 ≤ j ≤ m }) = L(V, W) and thus the required result follows. _

Before proceeding further, recall the following definition about a function.

PICT PICT DRAFT

Definition 4.2.9. [Inverse of a Function] Let f : S → T be any function.

1.
Then, a function g : T → S is called a left inverse of f if (g ∘ f)(x) = x, for all x ∈ S. That is, g ∘ f = Id, the identity function on S.
2.
Then, a function h : T → S is called a right inverse of f if (f ∘ h)(y) = y, for all y ∈ T. That is, f ∘ h = Id, the identity function on T.
3.
Then f is said to be invertible if it has a right inverse and a left inverse.

Remark 4.2.10. Let f : S → T be invertible. Then, it can be easily shown that any right inverse and any left inverse are the same. Thus, the inverse function is unique and is denoted by f-1. It is well known that f is invertible if and only if f is both one-one and onto.

Lemma 4.2.11. Let V and W be vector spaces over F and let T ∈L(V, W). If T is one-one and onto then, the map T-1 : W → V is also a linear transformation. The map T-1 is called the inverse linear transform of T and is defined by T-1(w) = v, whenever T(v) = w.

Proof. Part 1: As T is one-one and onto, by Theorem 4.2.3, dim(V) = dim(W). So, by Corollary 4.2.4, for each w ∈ W there exists a unique v ∈ V such that T(v) = w. Thus, one defines T-1(w) = v.

We need to show that T-1(α1w1 + α2w2) = α1T-1(w1) + α2T-1(w2), for all α1,α2 ∈ F and w1,w2 ∈ W. Note that by previous paragraph, there exist unique vectors v1,v2 ∈ V such that T-1(w1) = v1 and T-1(w2) = v2. Or equivalently, T(v1) = w1 and T(v2) = w2. So, T(α1v1 + α2v2) = α1w1 + α2w2, for all α1,α2 ∈ F. Hence, for all α1,α2 ∈ F, we get PICT PICT DRAFT

T- 1(α1w1 + α2w2 ) = α1v1 + α2v2 = α1T - 1(w1 )+ α2T -1(w2 ).
Thus, the required result follows. _

Example 4.2.12.

1.
Let T : ℝ2 → ℝ2 be given by (x,y) ⇝ (x + y,x - y). Then, verify that T-1 is given by ⇝( ) x+2y, x-2y.
2.
Let T ∈L(ℝn, ℝ[x;n - 1]) be given by (a1,…,an) ⇝∑ i=1naixi-1, for (a1,…,an) ∈ ℝn. Then, T-1 maps ∑ i=1naixi-1 ⇝ (a1,…,an), for each polynomial ∑ i=1naixi-1 ∈ ℝ[x;n- 1]. Verify that T-1 ∈L(ℝ[x;n - 1], ℝn).

Definition 4.2.13. [Singular, Non-singular Transformations] Let V and W be vector spaces over F and let T ∈L(V, W). Then, T is said to be singular if 0 ⊊ Ker(T). That is, Ker(T) contains a non-zero vector. If Ker(T) = {0} then, T is called non-singular.

Example 4.2.14. Let T ∈L(ℝ2, ℝ3) be defined by T( [ ]) x y = ⌊ ⌋ | x| ⌈ y⌉ 0. Then, verify that T is non-singular. Is T invertible? PICT PICT DRAFT

We now prove a result that relates non-singularity with linear independence.

Theorem 4.2.15. Let V and W be vector spaces over F and let T ∈L(V, W). Then the following statements are equivalent.

1.
T is one-one.
2.
T is non-singular.
3.
Whenever S ⊆ V is linearly independent then T(S) is necessarily linearly independent.

Proof. 1⇒2 Let T be singular. Then, there exists v≠0 such that T(v) = 0 = T(0). This implies that T is not one-one, a contradiction.

2⇒3 Let S ⊆ V be linearly independent. Let if possible T(S) be linearly dependent. Then, there exists v1,…,vk ∈ S and α = (α1,…,αk)T ≠0 such that ∑ i=1kαiT(vi) = 0. Thus, T( k ) ∑ αivi i=1 = 0. But T is nonsingular and hence we get ∑ i=1kαivi = 0 with α≠0, a contradiction to S being a linearly independent set.

3⇒1 Suppose that T is not one-one. Then, there exists x,y ∈ V such that x≠y but T(x) = T(y). Thus, we have obtained S = {x - y}, a linearly independent subset of V with T(S) = {0}, a linearly dependent set. A contradiction to our assumption. Thus, the required result follows. _

Definition 4.2.16. [Isomorphism of Vector Spaces] Let V and W be two vector spaces over F and let T ∈L(V, W). Then, T is said to be an isomorphism if T is one-one and onto. The vector spaces V and W are said to be isomorphic, denoted V~ =W, if there is an isomorphism from V to W.

We now give a formal proof of the statement in Remark 3.6.9.

PICT PICT DRAFT Theorem 4.2.17. Let V be an n-dimensional vector space over F. Then V~=Fn.

Proof. Let {v1,…,vn} be a basis of V and {e1,…,en}, the standard basis of Fn. Now define T(vi) = ei, for 1 ≤ i ≤ n and T( ) ∑n i=1αivi = ∑ i=1nαiei, for α1,…,αn ∈ F. Then, it is easy to observe that T ∈L(V, Fn), T is one-one and onto. Hence, T is an isomorphism. _

As a direct application using the countability argument, one obtains the following result

Corollary 4.2.18. The vector space ℝ over ℚ is not finite dimensional. Similarly, the vector space ℂ over ℚ is not finite dimensional.

We now summarize the different definitions related with a linear operator on a finite dimensional vector space. The proof basically uses the rank-nullity theorem and they appear in some form in previous results. Hence, we leave the proof for the reader.

Theorem 4.2.19. Let V be a vector space over F with dim V = n. Then the following statements are equivalent for T ∈L(V).

1.
T is one-one.
2.
Ker(T) = {0}.
3.
Rank(T) = n.
4.
T is onto.
5.
T is an isomorphism.
6.
If {v1,…,vn} is a basis for V then so is {T(v1),…,T(vn)}.
7.
T is non-singular. PICT PICT DRAFT
8.
T is invertible.

Exercise 4.2.20. Let V and W be two vector spaces over F and let T ∈L(V, W). If dim(V) is finite then prove that

1.
T cannot be onto if dim(V) < dim(W).
2.
T cannot be one-one if dim(V) > dim(W).

4.3 Matrix of a linear transformation

In Example 4.1.3.8, we saw that for each A ∈ Mm×n(ℂ) there exists a linear transformation T ∈L(ℂn, ℂm) given by T(x) = Ax, for each x ∈ ℂn. In this section, we prove that if V and W are vector spaces over F with dimensions n and m, respectively, then any T ∈L(V, W) corresponds to a set of m×n matrices. Before proceeding further, the readers should recall the results on ordered basis (see Section 3.6).

So, let A = (v1,...,vn ) and B = (w1,...,wm ) be ordered bases of V and W, respectively. Also, let A = [v1,…,vn] and B = [w1,…,wm] be the basis matrix of A and B, respectively. Then, using Equation (3.6.1), v = A[v]A and w = B[w]B, for all v ∈ V and w ∈ W. Thus, using the above discussion and Equation (4.1.1), we see that for T ∈L(V, W) and v ∈ V,

[ ] B [T (v)]B = T (v) = T (A[v]A) = T(A )[v]A = T (v ) ⋅⋅⋅ T(v ) [v]A [ ] 1[ n ] = B [T (v )] ⋅⋅⋅ B [T (v )] [v]A = B [T (v )] ⋅⋅⋅ [T (v )] [v]A. 1 B n B 1 B n B <img src= PICT DRAFT " class="math-display" >

Therefore, [T(v)]B = [[T(v1)]B,...,[T (vn )]B][v]A as a vector in W has a unique expansion in terms of basis elements. Note that the matrix [ ] [T (v1)]B ⋅⋅⋅ [T (vn )]B, denoted T[A,B], is an m×n matrix and is unique with respect to the ordered basis B as the i-th column equals [T(vi)]B, for 1 ≤ i ≤ n. So, we immediately have the following definition and result.

Definition 4.3.1. [Matrix of a Linear Transformation] Let A = (v1,…,vn) and B = (w1,…,wm) be ordered bases of V and W, respectively. If T ∈L(V, W) then the matrix T[A,B] is called the coordinate matrix of T or the matrix of the linear transformation T with respect to the basis A and B, respectively. When there is no mention of bases, we take it to be the standard ordered bases and denote the corresponding matrix by [T].

Note that if c is the coordinate vector of an element v ∈ V then, T[A,B]c is the coordinate vector of T(v). That is, the matrix T[A,B] takes coordinate vector of the domain points to the coordinate vector of its images.

Theorem 4.3.2. Let A = (v1,…,vn) and B = (w1,…,wm) be ordered bases of V and W, respectively. If T ∈L(V, W) then there exists a matrix S ∈ Mm×n(F) with

[ ] S = T[A,B ] = [T(v1)]B ⋅⋅⋅ [T(vn)]B and [T (x)]B = S [x ]A, for all x ∈ V.

PICT PICT DRAFT Remark 4.3.3. Let V and W be vector spaces over F with ordered bases A1 = (v1,…,vn) and B1 = (w1,…,wm), respectively. Also, for α ∈ F with α≠0, let A2 = (αv1,…,αvn) and B1 = (αw1,…,αwm) be another set of ordered bases of V and W, respectively. Then, for any T ∈L(V, W)

[ ] [ ] T[A2,B2 ] = [T (αv1)]B2 ⋅⋅⋅ [T (αv1)]B2 = [T(vn )]B1 ⋅⋅⋅ [T(v1)]B1 = T [A1, B1].
Thus, we see that the same matrix can be the matrix representation of T for two different pairs of bases.

We now give a few examples to understand the above discussion and Theorem 4.3.2.

PICT PICT DRAFT PICT PICT DRAFT

PIC


Figure 4.1: Counter-clockwise Rotation by an angle θ
PICT PICT DRAFT

Example 4.3.4.

1.
Let T ∈L(ℝ2) represent a counterclockwise rotation by an angle θ,0 ≤ θ < 2π. Then, using Figure 4.1, x = OP cosα and y = OP sinα, verify that
[ ] [ ] [ ] [ ][ ] x′ OP ′cos(α + θ) OP (cosα cosθ - sinα sin θ) cosθ - sin θ x y′ = OP ′sin(α + θ) = OP (sin αcos θ + cosα sin θ) = sinθ cosθ y .
Or equivalently, the matrix in the standard ordered basis of ℝ2 equals
[ ] [ ] [T ] = T (e1),T (e2) = cos θ - sinθ . sinθ cos θ
(4.3.1)

2.
Let T ∈L(ℝ2) with T((x,y)T ) = (x + y,x - y)T .
(a)
Then [T] = [ ] [T(e1)] [T (e2)] = [ ] 1 1 1 - 1. PICT PICT DRAFT
(b)
On the image space take the ordered basis B = ( [ ] [ ] ) 1 1 0 , 1. Then
[T] = [ ] [T (e )] [T(e )] 1 B 2 B = [[ ] [ ] ] 1 1 1 - 1 B B = [ ] 0 2 1 - 1.
(c)
In the above, let the ordered basis of the domain space be A = ([ ] [ ]) - 1 3 1 , 1. Then T[A,B] = [[ [ ]] [ [ ]] ] - 1 3 T 1 T 1 B B = [[ ] [ ] ] 0 4 - 2 2 B B = [ ] 2 2 - 2 2.
3.
Let A = (e1,e2) and B = (e1 + e2,e1 -e2) be two ordered bases of ℝ2. Then Compute T[A,A] and T[B,B], where T((x,y)T ) = (x + y,x - 2y)T .
Solution: Let A = Id2 and B = [ ] 1 1 1 - 1. Then, A-1 = Id2 and B-1 = 1- 2[ ] 1 1 1 - 1. So,
T[A,A] = [[ ([ ]) ] [ ([ ] )] ] 1 0 T , T 0 A 1 A = [[ ] [ ] ] 1 1 , 1 A - 2 A = [ ] 1 1 1 - 2 and
T[B,B] = [[ ([ ]) ] [ ( [ ]) ] ] 1 1 T , T 1 B - 1 B = [[ ] [ ] ] 2 0 , - 1 B 3 B = [ ] 1 3 23 23 2 - 2

as [ ] 2 - 1B = B-1[ ] 2 - 1 and [ ] 0 3B = B-1[ ] 0 3. Also, verify that T[B,B] = B-1T[A,A]B.

4.
Let T ∈L(ℝ3, ℝ2) be defined by T((x,y,z)T ) = (x + y - z,x + z)T . Determine [T].
By definition PICT PICT DRAFT
[[ ] [ ] [ ]] [ ] 1 1 - 1 1 1 - 1 [T ] = [[T (e1)],[T (e2)],[T(e3)]] = 1 , 0 , 1 = 1 0 1 .
5.
Define T ∈L(ℂ3) by T(x) = x, for all x ∈ ℂ3. Determine the coordinate matrix with respect to the ordered basis A = (e1,e2,e3) and B = ((1,0,0),(1,1,0),(1,1,1)).
By definition, verify that
⌊ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌋ ⌊ ⌋ 1 0 0 1 - 1 0 T[A,B ] = [[T (e )] ,[T(e )] ,[T (e )] ] = |⌈ |⌈0|⌉ ,|⌈1|⌉ ,|⌈0|⌉ |⌉ = |⌈0 1 - 1|⌉ 1 B 2 B 3 B 0 B 0 B 1 B 0 0 1
and
⌊⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌋ ⌊ ⌋ || 1| |1| |1| | | 1 1 1| T[B,A ] = ⌈⌈ 0⌉ ,⌈1⌉ ,⌈1⌉ ⌉ = ⌈ 0 1 1⌉ . 0 0 1 0 0 1 A A A
Thus, verify that T[B,A]-1 = T[A,B] and T[A,A] = T[B,B] = I3 as the given map is indeed the identity map.
6.
Fix S ∈ Mn(ℂ) and define T ∈L(ℂn) by T(x) = Sx, for all x ∈ ℂn. If A is the standard basis of ℂn then [T] = S as PICT PICT DRAFT
[T][:,i] = [T(ei)]A = [S(ei)]A = [S[:,i]]A = S [:,i], for 1 ≤ i ≤ n.
7.
Fix S ∈ Mm,n(ℂ) and define T ∈L(ℂn, ℂm) by T(x) = Sx, for all x ∈ ℂn. Let A and B be the standard ordered bases of ℂn and ℂm, respectively. Then T[A,B] = S as
(T [A,B ])[:,i] = [T(e )] = [S(e )] = [S[:,i]] = S[:,i], for 1 ≤ i ≤ n. i B i B B
8.
Fix S ∈ Mn(ℂ) and define T ∈L(ℂn) by T(x) = Sx, for all x ∈ ℂn. Let A = (v1,...,vn) and B = (u1,...,un) be two ordered basses of ℂn with respective basis matrices A and B. Then
[ ] [ ] T [A, B] = [T(v )] ⋅⋅⋅ [T(v )] = B-1T (v ) ⋅⋅⋅ B -1T(v ) [ 1 B 1 B] [ 1 ] 1 = B -1Sv1 ⋅⋅⋅ B -1Sv1 = B -1S v1 ⋅⋅⋅ vn = B -1SA.
In particular, if
(a)
A = B then T[A,A] = A-1SA. Thus, if S = In so that T = Id then Id[A,A] = In. PICT PICT DRAFT
(b)
S = In so that T = Id then Id[A,B] = B-1A, an invertible matrix. Similarly, Id[B,A] = A-1B. So, Id[B,A] ⋅Id[A,B] = (A-1B)(B-1A) = In.
9.
Let T( ) (x,y)T = (x + y,x-y)T and A = (e1,e1 + e2) be the ordered basis of ℝ2. Then, using Example 4.3.4.8a we obtain
[ ][ ] [ ] ( [ ])- 1[ ] 1 - 1 1 2 0 2 T[A,A ] = e1 e1 + e2 T(e1) T(e1 + e2) = = . 0 1 1 0 1 0

Example 4.3.5. [Finding T from T[A,B]]

1.
Let V and W be vector spaces over F with ordered bases A and B, respectively. Suppose we are given the matrix S = T[A,B]. Then determine the corresponding T ∈L(V, W).

Solution: Let B be the basis matrix corresponding to the ordered basis B. Then, using Equation (3.6.1) and Theorem 4.3.2, we see that

T(v) = B [T (v )]B = BT [A, B][v]A = BS [v]A.
2.
In particular, if V = W = Fn and A = B then we see that
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T (v ) = BSB -1v.
(4.3.2)

Exercise 4.3.6.

1.
Let T ∈L(ℝ2) represent the reflection about the line y = mx. Find [T].
2.
Let T ∈L(ℝ3) represent the reflection about the X-axis. Find [T].
3.
Let T ∈L(ℝ3) represent the counterclockwise rotation around the positive Z-axis by an angle θ,0 ≤ θ < 2π. Find its matrix with respect to the standard ordered basis of ℝ3. [Hint: Is ⌊ ⌋ cosθ - sin θ 0 | | ⌈ sin θ cosθ 0⌉ 0 0 1 the required matrix?]
4.
Define a function D ∈L(ℝ[x;n]) by D(f(x)) = f′(x). Find the matrix of D with respect to the standard ordered basis of ℝ[x;n]. Observe that Rng(D) ⊆ ℝ[x;n - 1].

4.4 Similarity of Matrices

Let V be a vector space over F with dim(V) = n and ordered basis B. Then any T ∈L(V) corresponds to a matrix in Mn(F). What happens if the ordered basis needs to change? We answer this in this subsection.

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Figure 4.2: Composition of Linear Transformations
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Theorem 4.4.1 (Composition of Linear Transformations). Let V, W and ℤ be finite dimensional vector spaces over F with ordered bases B,C and D, respectively. Also, let T ∈ L(V, W) and S ∈L(W, ℤ). Then S ∘ T = ST ∈L(V, ℤ) (see Figure 4.2). Then

(ST )[B,D ] = S[C,D ]⋅T[B,C].

Proof. Let B = (u1,…,un),C = (v1,…,vm) and D = (w1,…,wp) be the ordered bases of V, W and ℤ, respectively. Then using Theorem 4.3.2, we have

(ST )[B, D] = [[ST(u1 )]D ,...,[ST (un)]D ] = [[S (T(u1))]D ,...,[S (T (un))]D ] = [S[C,D ][T (u )] ,...,S [C,D ][T(u )] ] 1 C n C = S [C, D][[T (u1)]C ,...,[T(un )]C] = S [C,D ]⋅T [B, C].
Hence, the proof of the theorem is complete. _

As an immediate corollary of Theorem 4.4.1 we have the following result.

PICT PICT DRAFT Theorem 4.4.2 (Inverse of a Linear Transformation). Let V is a vector space with dim(V) = n. If T ∈L(V) is invertible then for any ordered basis B and C of the domain and co-domain, respectively, one has (T[C,B])-1 = T-1[B,C]. That is, the inverse of the coordinate matrix of T is the coordinate matrix of the inverse linear transform.

Proof. As T is invertible, TT-1 = Id. Thus, Example 4.3.4.8a and Theorem 4.4.1 imply

- 1 - 1 In = Id[B, B] = (T T )[B,B ] = T [C,B ]⋅T [B,C ].
Hence, by definition of inverse, T-1[B,C] = (T [C,B])-1 and the required result follows. _

Exercise 4.4.3. Find the matrix of the linear transformations given below.

1.
Let B = (x1,x2,x3) be an ordered basis of ℝ3. Now, define T ∈L(ℝ3) by T(x1) = x2, T(x2) = x3 and T(x3) = x1. Determine T[B,B]. Is T invertible?
2.
Let B = (1,x,x2,x3) be an ordered basis of ℝ[x;3] and define T ∈L(ℝ[x;3]) by T(1) = 1, T(x) = 1 + x, T(x2) = (1 + x)2 and T(x3) = (1 + x)3. Prove that T is invertible. Also, find T[B,B] and T-1[B,B].

Let V be a finite dimensional vector space. Then, the next result answers the question “what happens to the matrix T[B,B] if the ordered basis B changes to C?”

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Figure 4.3: T[C,C] = Id[B,C] ⋅ T[B,B] ⋅ Id[C,B] - Similarity of Matrices
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Theorem 4.4.4. Let B = (u1,…,un) and C = (v1,…,vn) be two ordered bases of V and Id the identity operator. Then, for any linear operator T ∈L(V)

T [C, C] = Id[B,C]⋅T [B, B]⋅Id[C,B] = (Id[C,B])-1 ⋅T[B,B]⋅ Id[C,B ].
(4.4.1)

Proof. As Id is an identity operator, T[B,C] as (Id ∘ T ∘ Id)[B,C] (see Figure 4.3 for clarity). Thus, using Theorem 4.4.1, we get

T[B,C] = (Id ∘T ∘ Id)[B,C ] = Id[B,C]⋅T [B, B]⋅Id[C,B].
Hence, using Theorem 4.4.2, the required result follows. _

Let V be a vector space and let T ∈L(V). If dim(V) = n then every ordered basis B of V gives an n × n matrix T[B,B]. So, as we change the ordered basis, the coordinate matrix of T changes. Theorem 4.4.4 tells us that all these matrices are related by an invertible matrix. Thus, we are led to the following definitions.

PICT PICT DRAFT Definition 4.4.5. [Change of Basis Matrix] Let V be a vector space with ordered bases B and C. If T ∈L(V) then, T[C,C] = Id[B,C] ⋅ T[B,B] ⋅ Id[C,B]. The matrix Id[B,C] is called the change of basis matrix (also, see Theorem 3.6.7) from B to C.

Definition 4.4.6. [Similar Matrices] Let X,Y ∈ Mn(ℂ). Then, X and Y are said to be similar if there exists a non-singular matrix P such that P-1XP = Y ⇔ XP = PY .

Example 4.4.7. Let B = (1 + x,1 + 2x + x2,2 + x) and C = (1,1 + x,1 + x + x2) be ordered bases of ℝ[x;2]. Then, verify that Id[B,C]-1 = Id[C,B], as

⌊- 1 1 - 2⌋ 2 | | Id[C,B ] = [[1]B,[1 + x]B,[1 + x + x ]B ] = ⌈ 0 0 1 ⌉ and 1 0 1 ⌊ ⌋ 0 - 1 1 Id[B, C] = [[1 + x]C,[1+ 2x+ x2]C,[2+ x]C] = |⌈ 1 1 1|⌉ . 0 1 0

PICT PICT DRAFT Exercise 4.4.8.

1.
Let V be a vector space with dim(V) = n. Let T ∈L(V) satisfy Tn-1≠0 but Tn = 0. Then, use Exercise 4.1.14.4 to get an ordered basis B = (u,T(u),…,Tn-1(u)) of V.
(a)
Now, prove that T[B,B] = ⌊ ⌋ 0 0 0 ⋅⋅⋅ 0 ||1 0 0 ⋅⋅⋅ 0|| || || ||0 1 0 ⋅⋅⋅ 0|| |... ... ... ...| ⌈ ⌉ 0 0 ⋅⋅⋅ 1 0.
(b)
Let A ∈ Mn(ℂ) satisfy An-1≠0 but An = 0. Then, prove that A is similar to the matrix given in Part 1a.
2.
Let A be an ordered basis of a vector space V over F with dim(V) = n. Then prove that the set of all possible matrix representations of T is given by (also see Definition 4.4.5)
{S ⋅T[A,A ]⋅S- 1|S ∈ Mn (F) is an invertible matrix}.
3.
Let B1(α,β) = {(x,y)T ∈ ℝ2 : (x - α)2 + (y - β)2 ≤ 1}. Then, can we get a linear transformation T ∈L(ℝ2) such that T(S) = W, where S and W are given below?
(a)
S = B1(0,0) and W = B1(1,1).
(b)
S = B1(0,0) and W = B1(.3,0).
(c)
S = B1(0,0) and W = hull(±e1,±e2), where hull means the convex hull.
(d)
S = B1(0,0) and W = {(x,y)T ∈ ℝ2 : x2 + y2∕4 = 1}.
(e)
S = hull(±e1,±e2) and W is the interior of a pentagon.
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4.
Let V, W be vector spaces over F with dim(V) = n and dim(W) = m and ordered bases B and C, respectively. Define IB,C : L(V, W) → Mm,n(F) by IB,C(T) = T[B,C]. Show that IB,C is an isomorphism. Thus, when bases are fixed, the number of m × n matrices is same as the number of linear transformations.__________________________________________________________
5.
Define T ∈L(ℝ3) by T((x,y,z)T ) = (x + y + 2z,x-y - 3z,2x + 3y + z)T . Let B be the standard ordered basis and C = ((1,1,1),(1,-1,1),(1,1,2)) be another ordered basis of ℝ3. Then find
(a)
matrices T[B,B] and T[C,C].
(b)
the matrix P such that P-1T[B,B]P = T[C,C].

4.5 Dual Space*

Definition 4.5.1. [linear Functional] Let V be a vector space over F. Then a map T ∈L(V, F) is called a linear functional on V.

Example 4.5.2.

1.
Let a ∈ ℂn be fixed. Then, T(x) = a*x is a linear function from ℂn to ℂ.
2.
Define T(A) = tr(A), for all A ∈ Mn(ℝ). Then, T is a linear functional from Mn(ℝ) to ℝ.
3.
Define T(f) = ∫ abf(t)dt, for all f ∈ C([a,b], ℝ). Then, T is a linear functional from L(C([a,b], ℝ) to ℝ. PICT PICT DRAFT
4.
Define T(f) = ∫ abt2f(t)dt, , for all f ∈C([a,b], ℝ). Then, T is a linear functional from L(C([a,b], ℝ) to ℝ.
5.
Define T : ℂ3 → ℂ by T((x,y,z)T ) = x. Is it a linear functional?
6.
Let B be a basis of a vector space V over F. For a fixed element u ∈B, define
{ 1 if x = u T (x) = 0 if x ∈ B \ u.
Now, extend T linearly to all of V. Does, T give rise to a linear functional?

Definition 4.5.3. [Dual Space] Let V be a vector space over F. Then L(V, F) is called the dual space of V and is denoted by V*. The double dual space of V, denoted V**, is the dual space of V*.

We first give an immediate corollary of Theorem 4.2.17.

Corollary 4.5.4. Let V and W be vector spaces over F with dim V = n and dim W = m.

1.
Then L(V, W)~ =Fmn. Moreover, {fij|1 ≤ i ≤ n,1 ≤ j ≤ m} is a basis of L(V, W).
2.
In particular, if W = F then L(V, F) = V*~ =Fn. Moreover, if {v1,…,vn} is a basis of V then the set {fi|1 ≤ i ≤ n} is a basis of V*, where fi(vk) = { 1, if k = i 0, k ⁄= i. The basis {fi|1 ≤ i ≤ n} is called the dual basis of Fn.
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Exercise 4.5.5. Let V be a vector space. Suppose there exists v ∈ V such that f(v) = 0, for all f ∈ V*. Then prove that v = 0.

So, we see that V* can be understood through a basis of V. Thus, one can understand V** again via a basis of V*. But, the question arises “can we understand it directly via the vector space V itself?” We answer this in affirmative by giving a canonical isomorphism from V to V**. To do so, for each v ∈ V, we define a map Lv : V*→ F by Lv(f) = f(v), for each f ∈ V*. Then Lv is a linear functional as

Lv(αf + g) = (αf + g )(v) = αf(v)+ g (v ) = αLv (f)+ Lv (g).
So, for each v ∈ V, we have obtained a linear functional Lv ∈ V**. Note that, if v≠w then, Lv≠Lw. Indeed, if Lv = Lw then, Lv(f) = Lw(f), for all f ∈ V*. Thus, f(v) = f(w), for all f ∈ V*. That is, f(v - w) = 0, for each f ∈ V*. Hence, using Exercise 4.5.5, we get v - w = 0, or equivalently, v = w.

We use the above argument to give the required canonical isomorphism.

Theorem 4.5.6. Let V be a vector space over F. If dim(V) = n then the canonical map T : V → V** defined by T(v) = Lv is an isomorphism.

Proof. Note that for each f ∈ V*,

PICT PICT DRAFT Lαv+u (f) = f(αv + u) = αf(v)+ f(u ) = αLv (f)+ Lu (f) = (αLv + Lu )(f).
Thus, Lαv+u = αLv + Lu. Hence, T(αv + u) = αT(v) + T(u). Thus, T is a linear transformation. For verifying T is one-one, assume that T(v) = T(u), for some u,v ∈ V. Then, Lv = Lu. Now, use the argument just before this theorem to get v = u. Therefore, T is one-one.

Thus, T gives an inclusion (one-one) map from V to V**. Further, applying Corollary 4.5.4.2 to V*, gives dim(V**) = dim(V*) = n. Hence, the required result follows. _

We now give a few immediate consequences of Theorem 4.5.6.

Corollary 4.5.7. Let V be a vector space of dimension n with basis B = {v1,…,vn}.

1.
Then, a basis of V**, the double dual of V, equals D = {Lv1,…,Lvn}. Thus, for each T ∈ V** there exists x ∈ V such that T(f) = f(x), for all f ∈ V*. Or equivalently, there exists x ∈ V such that T = T

x.IfC = {f˙1, …, f˙n}isthedualbasisofVˆ*definedusingthebasisB(seeCorollary 4.5.4.2)thenDisindeedthedualbasisofVˆ**obtainedusingthebasisCofVˆ*.Thus,eachbasisofVˆ*isthedualbasisofsomebasisofV.

Proof. Part 1 is direct as T : V → V** was a canonical inclusion map. For Part 2, we need to show that

{ { 1, if j = i 1, if j = i Lvi (fj) = 0, if j ⁄= i or equivalently fj(vi) = 0, if j ⁄= i
which indeed holds true using Corollary 4.5.4.2. _

Let V be a finite dimensional vector space. Then Corollary 4.5.7 implies that the spaces V and V* are naturally dual to each other.

We are now ready to prove the main result of this subsection. To start with, let V and W be vector spaces over F. Then, for each T ∈L(V, W), we want to define a map ^ T : W*→ V*. So, if g ∈ W* then, T^(g) a linear functional from V to F. So, we need to be evaluate ^T(g) at an element of V. Thus, we define ( ) ^T(g)(v) = g(T(v)), for all v ∈ V. Now, we note that ^T ∈L(W*, V*), as for every PICT PICT DRAFT g,h ∈ W*,

( ) ( ) ^T(αg + h ) (v) = (αg + h)(T (v )) = αg (T(v)) + h(T (v )) = α ^T(g) + ^T(h) (v),
for all v ∈ V implies that ^T(αg + h) = α^T(g) + T^(h).

2. Theorem 4.5.8. Let V and W be two vector spaces over F with ordered bases A = (v1,…,vn) and B = (w1,…,wm), respectively. Also, let A* = (f1,…,fn) and B* = (g1,…,gm) be the corresponding ordered bases of the dual spaces V* and W*, respectively. Then,

^T [B *,A *] = (T [A, B])T ,
the transpose of the coordinate matrix T.

Proof. Note that we need to compute ^T[B*,A*] = [[ ] [ ] ] T^(g1) ,..., ^T(gm ) A* A* and prove that it equals the transpose of the matrix T[A,B]. So, let

⌊ ⌋ a11 a12 ⋅⋅⋅ a1n || a21 a22 ⋅⋅⋅ a2n|| T [A, B] = [[T(v1)]B,...,[T (vn)]B ] = || . . . . ||. ⌈ .. .. .. .. ⌉ a a ⋅⋅⋅ a m1 m2 mn
Thus, to prove the required result, we need to show that
⌊ ⌋ | aj1| [ ] | aj2| ∑n T^(gj) A* = [f1,...,fn]|| .. || = ajkfk, for 1 ≤ j ≤ m. ⌈ . ⌉ k=1 ajn
(4.5.1)

Now, recall that the functionals fi’s and gj’s satisfy ( ∑n ) αkfk k=1(vt) = ∑ k=1nαk(fk(vt)) = αt, for 1 ≤ t ≤ n and [gj(w1 ),...,gj(wm )] = ejT , a row vector with 1 at the j-th place and 0, elsewhere. So, let B = [w1,...,wm ] and evaluate ^T(gj) at vt’s, the elements of A.

( ) ^T (gj) (vt) = gj (T (vt)) = gj (B [T(vt)]B) = [gj(w1 ),...,gj (wm )][T(vt)]B ⌊ ⌋ a1t ( ) || a2t|| ∑n = eTj|| . || = ajt = ajkfk (vt). ⌈ .. ⌉ k=1 amt
Thus, the linear functional T^(gj) and ∑ k=1najkfk are equal at vt, for 1 ≤ t ≤ n, the basis vectors of V. Hence T^(gj) = ∑ k=1najkfk which gives Equation (4.5.1). _ PICT PICT DRAFT

Remark 4.5.9. The proof of Theorem 4.5.8 also shows the following.

1.
For each T ∈L(V, W) there exists a unique map T^ ∈L(W*, V*) such that
( ^ ) * T(g) (v ) = g (T(v)), for each g ∈ W .
2.
The coordinate matrices T[A,B] and ^T[B*,A*] are transpose of each other, where the ordered bases A* of V* and B* of W* correspond, respectively, to the ordered bases A of V and B of W.
3.
Thus, the results on matrices and its transpose can be re-written in the language a vector space and its dual space.

4.6 Summary

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