Problem Solving - Linear Transformation

Problem Solving - Linear Transformation  

M.Velrajan

 We continue our discussion on Problem Solving - Pure Mathematics with elementary exercises on linear transformations given in Topics in Algebra, 

by I. N. Herstein, second edition.

Let V be a finite dimensional vector space over a field F and 

A(V) = Hom (V, V) be the algebra of all linear transformations of V into V.

1. S ∈ A(V) is regular if and only if whenever v1, . . . , vn ∈ V are linearly independent, then v1S, . . . , vnS are also linearly independent.

Suppose S ∈ A(V) is regular. 

Let v1, . . . , vn ∈ V be linearly independent. 

 To prove v1S, . . . , vnS are linearly independent we prove 

𝛂1v1 S + . . . + 𝛂nvn S = 0 ⟹  𝛂1 = . . . = 𝛂n = 0.

Then 𝛂1v1 S + . . . + 𝛂nvn S = 0 

         ⟹ (𝛂1v1 + . . . + 𝛂nvn) S = 0 (since S is linear)  

         ⟹ 𝛂1v1 + . . . + 𝛂nvn = 0 (since S is one-one, 

                                                    vS = 0 = 0S ⟹ v = 0) 

         ⟹ 𝛂1 = . . . = 𝛂n = 0.

Hence v1S, . . . , vnS are linearly independent.

Note that we have used only S is one-one, not S is regular and 

V is finite dimensional. 

Conversely suppose whenever v1, . . . , vn ∈ V are linearly independent, 

then v1S, . . . , vnS are also linearly independent.

Since V is finite dimensional, 

S ∈ A(V) is regular if and only if S ∈ A(V) is a bijection 

                                  if and only if S is an injection. 

So, it is enough to prove S is one-one.

But from the converse hypothesis can we show directly that S is one-one?

No, so we prove S is one-one contrapositively.

Claim : S is one-one

Suppose S is not one-one.

Then there exist v1 ≠ v2 ∈ V such that v1S = v2S 

i.e. there exists v =  v1 - v2 ≠ 0 ∈ V such that vS = (v1 - v2)S = 0.

Since v ≠ 0, v is linearly independent and 

since vS = 0, vS is not linearly independent. 

This contradicts the hypothesis whenever v1, . . . , vn ∈ V are linearly independent, then v1S, . . . , vnS are also linearly independent.

Hence S is one-one.

Observe that we have proved that

S ∈ A(V) is one-one if and only if whenever v1, . . . , vn ∈ V are linearly independent, then v1S, . . . , vnS are also linearly independent.

Now, S ∈ A(V) is regular if and only if S is a bijection.

Since V is finite dimensional, 

S ∈ A(V) is a bijection if and only if S is an injection 

                                      if and only if S is a surjection.

Hence, S ∈ A(V) is regular if and only if whenever v1, . . . , vn ∈ V are linearly independent, then v1S, . . . , vnS are also linearly independent.

2. T ∈ A(V) is completely determined by its values on a basis of V.

Let {v1, . . . , vn} be a basis of V.

Let T ∈ A(V) and wi = viT, i = 1, 2, . . . , n.

Then w1, . . . , wn are not necessarily distinct elements of V.

Let v ∈ V and v = 𝛂1v1 + . . . + 𝛂nvn.

Then vT = (𝛂1v1 + . . . + 𝛂nvn)T 

                = 𝛂1v1T + . . . + 𝛂nvnT 

                = 𝛂1w1 + . . . + 𝛂nwn.

Hence T is determined by, 

wi = viT, i = 1, 2, . . . , n, its values on a basis of V.

Suppose S ∈ A(V) and  viS = wi = viT, i = 1, 2, . . . , n.

Let v ∈ V and v = 𝛂1v1 + . . . + 𝛂nvn. 

Then vS = (𝛂1v1 + . . . + 𝛂nvn)S 

               = 𝛂1v1S + . . . + 𝛂nvnS 

               = 𝛂1v1T + . . . + 𝛂nvnT 

               =  (𝛂1v1 + . . . + 𝛂nvn)T = vT.

Hence S = T.

Therefore, 

T is completely determined by its values on a basis of V.

2.1. Observation :

If {v1, . . . , vn} is a basis of V and 

w1, . . . , wn are, not necessarily distinct, elements of V. 

Then there exists an unique T ∈ A(V) such that viT = wi, i = 1, 2, . . . , n.

For, 

Let v ∈ V and v = 𝛂1v1 + . . . + 𝛂nvn. 

Define  

vT = (𝛂1v1 + . . . + 𝛂nvn)T = 𝛂1w1 + . . . + 𝛂nwn.

Then T ∈ A(V) and viT = wi, i = 1, 2, . . . , n. (verify)

Since T ∈ A(V) is completely determined by its values on a basis of V, 

the uniqueness of T follows.

3. If dimF V > 1 then A(V) is not commutative.

To prove A(V) is not commutative, 

we have to prove there are S, T ∈ A(V) such that ST ≠ TS 

i.e. vST ≠ vTS, for an element v ∈ V.

Think how to get such S, T ∈ A(V) and v ∈ V using dimF V > 1 

i.e using a basis (having more than one element) of V. 

Yes, we can define S, T ∈ A(V) using the previous observation.

But we know only the basis elements of V, so we have to take 

the wi s required to define S, T only from the basis elements of V 

and the zero of V and vST ≠ vTS, for a basis element v. 

Since dimF V may be 2, we have to use only 2 of the basis elements as wi s.

So we use one of the basis elements and 0 as wi s to define S 

and the two of the basis elements and 0 as wi s to define T 

and vST ≠ vTS, for one of the basis elements v.  

Let dimF V = n > 1 and {v1, . . . , vn} be a basis of V.

Define S, T ∈ A(V) by 

v1S = v1 and viS = 0, i ≠ 1 and 

v1T = v2 and viT = 0, i ≠ 1.

Then v1ST = (v1S)T = v1T = v2 and

v1TS = (v1T)S = v2S = 0.

Hence v1ST ≠ v1TS and hence ST ≠ TS.

So, A(V) is not commutative.

Think : Where dimF V > 1 is used ?

When dimF V = 1, what is A(V) ?

 

4. In A(V), let Z = {T ∈ A(V) / ST = TS for all S ∈ A(V)}. Then Z merely consists of the multiples of the unit element of A(V) by the elements of F.  

We have to prove that Z = {𝛂I / 𝛂 ∈ F}. 

Clearly S(𝛂I) = 𝛂(IS), for all S ∈ A(V).

So, we have to prove that 

if ST = TS for all S ∈ A(V) then T = 𝛂I, for an 𝛂 ∈ F.

We can use ST = TS for all S ∈ A(V) only if we know some S ∈ A(V).

Think how to get S ∈ A(V) using dimF V is finite alone. 

Yes, if dimF V = n then dimF A(V) = n2, should come to mind. 

But how does it give S ∈ A(V). 

If we have learned properly, immediately after learning the proof of 

“ if dimF V = n then dimF A(V) = n2 ” 

we should have observed that 

if  {v1, . . . , vn} is a basis of V then {Tij / i, j = 1, 2, . . . , n}, 

where Tij is defined by

vkTij = vj, if k = i and vkTij = 0, k ≠ i, is a basis of A(V). 

And in fact if T ∈ A(V) and 

 

Oh, we can use Tij s as S 

i.e. TrsT = TTrs for all r, s = 1, 2, . . . , n.

So we need TijTrs and the relation between I and Tij.

We also observe that the matrix of T = (𝛂ij) and 

the matrix of Tij is the n × n matrix whose entries are all 0, 

except 1 in the (i, j)th entry. 

By properties of matrices TijTrs = Tis, if j = r and TijTrs = 0, j ≠ r. 

Let {v1, . . . , vn} be a basis of V.

Then {Tij / i, j = 1, 2, . . . , n}, where Tij is defined by

vkTij = vj, if k = i and vkTij = 0, k ≠ i, is a basis of A(V).  

Let T ∈ Z. Then ST = TS for all S ∈ A(V).

In particular, TrsT = TTrs for all r, s = 1, 2, . . . , n.

 

i.e. vsT = 𝛂vs = 𝛂vsI = vs(𝛂I), for all s = 1, 2, . . . , n, 

where 𝛂 = 𝛂11 = 𝛂22 = . . . = 𝛂nn and I is the unit element of A(V).

Since any T ∈ A(V) is completely determined by its values on a basis of V,

T = 𝛂I.

Also for any 𝛂 ∈ F, 𝛂I ∈ Z.

Therefore, Z = {𝛂I / 𝛂 ∈ F}.

Aliter : 

Let {v1, . . . , vn} be a basis of V.

Then {Tij / i, j = 1, 2, . . . , n}, where Tij is defined by

vkTij = vj, if k = i and vkTij = 0, k ≠ i, is a basis of A(V).  

Let T ∈ Z. Then ST = TS for all S ∈ A(V).

Let viT = 𝛂i1v1 + . . . + 𝛂invn, i = 1, 2, . . . , n.

Then viT = (viTii)T = viTiiT = viTTii = (viT)Tii 

                = (𝛂i1v1 + . . . + 𝛂invn)Tii  

                = 𝛂i1v1Tii + . . . + 𝛂invnTii = 𝛂iivi.

And viTTij = (viT)Tij = (𝛂iivi)Tij = 𝛂ii(viTij) = 𝛂iivj

Also viTTij = viTijT = (viTij)T = vjT = 𝛂jjvj.

So, 𝛂iivj  = 𝛂jjvj

i.e. (𝛂ii - 𝛂jj)vj = 0

Since vj ≠ 0, 𝛂ii - 𝛂jj = 0

i.e. 𝛂ii = 𝛂jj 

Hence viT = 𝛂vi, i = 1, 2, . . . , n, 𝛂 = 𝛂ii = 𝛂jj.

i.e. viT = 𝛂(viI) = vi(𝛂I), i = 1, 2, . . . , n,  and I is the unit element of A(V).

Hence T = 𝛂I.

Also for any 𝛂 ∈ F, 𝛂I ∈ Z.

Therefore, Z = {𝛂I / 𝛂 ∈ F}.

5. If dimF V > 1 then A(V) has no two-sided ideals other than (0) and A(V).

We have to prove that, if K ≠ (0) is a two-sided ideal of A(V) 

then K = A(V).

K = A(V) if and only if K has an unit of A(V).

Known units of A(V) are 𝛂I, 𝛂 ≠ 0 ∈ F and I is the unit element of A(V).

Think how to get 𝛂I, 𝛂 ≠ 0 ∈ F from T and using 

K is a two-sided ideal of A(V).

Since K ≠ (0) there is a T ≠ 0 ∈ K and hence ST, TS ∈ A(V) for all S ∈ A(V).

What is the known S ∈ A(V) ?

dimF V > 1 is given. So we have a basis of V and 

hence the basis Tij of A(V) is known. 

We have to use T ≠ 0 and Tij of A(V). 

So express T in terms of Tij.  

Let {v1, . . . , vn} be a basis of V.

Then {Tij / i, j = 1, 2, . . . , n}, where Tij is defined by

vkTij = vj, if k = i and vkTij = 0, k ≠ i, is a basis of A(V).  

Let K ≠ (0) be a two-sided ideal of A(V) and T ≠ 0 ∈ K.

 

Since 𝛂rs ≠ 0, 𝛂rsI is an unit in A(V).

Hence, K = A(V).

Therefore, (0) and A(V) are the only two-sided ideals of A(V).

Think : Where dimF V > 1 is used ?

When dimF V = 1, what are the only two-sided ideals of A(V) ?

6. If F is the field of integers modulo 2 and if V is two dimensional over F then compute the group of regular elements in A(V) and prove that this group is isomorphic to S3, the symmetric group of degree 3.

Let {v1, v2} be the basis of V over the field F = {0, 1}.

Since F is the field of integers modulo 2, 

2v = v + v = 0 for all v ∈ V.

Hence V = {0, v1, v2, v1 + v2} 

                = {0, v1, v2, v3}, where v3 = v1 + v2.

Any T ∈ A(V) is completely determined by its values on a basis of V 

i.e. by the values of  v1T and v2T.

T ∈ A(V) is regular if and only if T maps V onto V 

if and only if  

T is a bijection from V onto V, 0T = 0 and vT ≠ 0, for all v ≠ 0 ∈ V 

if and only if 

0T = 0 and u1 = v1T, u2 = v2T, u3 = v3T are v1, v2, v3 in some order

if and only if  

0T = 0 and T| {v1, v2, v3} is a bijection

The group G of regular elements in A(V) consists of

T1 defined by v1T1 = v1 and v2T1 = v2  (v3T1 = v3); 

T2 defined by v1T2 = v2 and v2T2 = v1  (v3T2 = v3);

T3 defined by v1T3 = v3 and v2T3 = v2  (v3T3 = v1);

T4 defined by v1T4 = v1 and v2T4 = v3  (v3T4 = v2);

T5 defined by v1T5 = v2 and v2T5 = v3  (v3T5 = v1);

T6 defined by v1T6 = v3 and v2T6 = v1  (v3T6 = v2). 

Since 0Ti = 0 and Ti | {v1, v2, v3} ∈ S3, and 

the operation on both G and S is the composition 

of maps G can be considered as S3. 

(or) since G is a non-abelian group of order 6 

and S3 is the only non-abelian group of order 6 (up to isomorphism), 

G is isomorphic to S3.

 

7. If E ∈ A(V) is an idempotent then V = V0 ⊕ V1, where v0E = 0 for all v0 ∈ V0 and v1E = v1 for all v1 ∈ V1.

Since E2 = E, E(I - E) = 0.

Let V0 = V(I - E) and V1 = VE. 

If v0 ∈ V0 then v0 = v(I - E) for some v ∈ V, hence 

v0E = v(I - E)E = v(E - E2) = v0 = 0.

If v1 ∈ V1 then v1 = vE for some v ∈ V, hence 

v1E = (vE)E = vE2 = vE = v1.

Now, V0 and V1 are subspaces of V (range of (I - E) and E) and

hence V0 + V1 is a subspace of V.

And if v ∈ V then v = v - vE + vE = v(I - E) + vE ∈ V0 + V1.

Hence V = V0 + V1.

If v ∈ V0 ∩ V1 then v ∈ V0 and v ∈ V1, 

hence vE = 0 and vE = v i.e. v = 0.

Hence V0 ∩ V1 = {0}.

Therefore, V = V0 ⊕ V1.

8. If T ∈ AF(V), F of characteristic not 2, satisfies T3 = T then V = V0 ⊕ V1 ⊕ V2, where v0T = 0 for all v0 ∈ V0, v1T = v1 for all v1 ∈ V1, and v2T = - v2 for all v2 ∈ V2.

Since T3 = T, T(I - T2) = 0. 

Let V0 = V(I - T2), V1 = VT(I + T) and V2 = VT(I - T).

If v0 ∈ V0 then v0 = v(I - T2) for some v ∈ V, hence 

v0T = v(I - T2)T = v(T - T3) = v0 = 0.

If v1 ∈ V1 then v1 = vT(I + T) for some v ∈ V, hence 

v1T = (vT(I + T))T = v(T2 + T3) = v(T2 + T) = vT(T + I) = vT(I + T) = v1.

If v2 ∈ V2 then v2 = vT(I - T) for some v ∈ V, hence 

v2T = (vT(I - T))T = v(T2 - T3) = v(T2 - T) 

       = vT(T - I) = v(-T(I - T)) = - vT(I - T) = - v2.

Now, V0, V1 and V2 are subspaces of V 

(range of (I - T2), T(I + T) and T(I - T)) 

and hence V0 + V1 + V2 is a subspace of V.

And if v ∈ V then 

v = v - vT2 + vT2 ∈ V(I - T2) + VT2 

                             ⊆ V(I - T2) + VT 

                             = V(I - T2) + VT(I + I) (verify)

                             = V(I - T2) + VT(I + T +  I - T)

                             ⊆  V(I - T2) + VT(I + T) + VT(I - T)

                             = V0 + V1 + V2. 

(Note that W(S + T) ⊆ WS + WT), but W(S + T) ≠  WS + WT)

Hence V = V0 + V1 + V2.

If v ∈ V0 ∩ V1 then v ∈ V0 and v ∈ V1, 

hence vT = 0 and vT = v i.e. v = 0.

Hence V0 ∩ V1 = {0}.

If v ∈ V0 ∩ V2 then v ∈ V0 and v ∈ V2, 

hence vT = 0 and vT = - v i.e. v = 0.

Hence V0 ∩ V2 = {0}.

If v ∈ V1 ∩ V2 then v ∈ V1 and v ∈ V2, 

hence vT = v and vT = - v 

i.e. 2v = 0 and hence v = 0, since the characteristic of F is not 2.

Hence V1 ∩ V2 = {0}.

Therefore, V = V0 ⊕ V1 ⊕ V2.

Let A be an algebra, with unit element, over F then A is isomorphic to a subalgebra of A(V) for some vector space V over F. (Lemma 6.1.1 in Topics in Algebra)

 9. Let A be an algebra, without unit element, over F then A is isomorphic to a subalgebra of A(V) for some vector space V over F.

By the above lemma it is enough to show that A is isomorphic to a subalgebra of an algebra B, with unit element, over F.

Let B = F × A.

Define (𝛂, x) + (𝛃, y) = (𝛂 + 𝛃, x + y)

(𝛂, x)(𝛃, y) = (𝛂𝛃, 𝛂y +𝛃x + xy)

and 𝛃(𝛂, x) = (𝛃𝛂, 𝛃x). 

Then B is an algebra, with unit element (1, 0), over F and 

x → (0, x) is an algebra isomorphism of A into B.

Hence A is isomorphic to a subalgebra of the algebra B, with unit element, over F.

So, we have : 

If A is an algebra, without unit element, over F then A is isomorphic to a subalgebra of an algebra B, with unit element, over F.

Since B is isomorphic to a subalgebra of A(V) for some vector space V over F,

A is isomorphic to a subalgebra of A(V) for some vector space V over F.

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