Abstract Algebra 21
THE DOUBLE DUAL SPACE (BIDUAL SPACE) and THE ADJOINT OF A LINEAR TRANSFORMATION
Objectives
From this session a learner is expected to achieve the following
1. Study that every basis for is the dual of some basis for .
2. Learn the concept of annihilator of the annihilator of a subset.
3. Learn the notion of hyperspace.
4. Understand the notion of maximal proper subspace and hyperspace.
Sections
1. Introduction
2. Dual of a Dual Space
3. Annihilator of the Annihilator of a Set
4. Hyperspaces
5. The Adjoint of a Linear Transformation
Introduction
In this session we discuss the relation between basis for a finite dimensional vector space V and the basis for its dual space V*. We shall answer a question about dual bases whether every basis for is the dual of some basis for . One way to answer this question is to consider , the dual space of . In this session along with the discussion of double dual space, the concept of annihilator of the annihilator of a subset and the notion of hyperspace shall be described. We shall also discuss the adjoint of a linear transformation.
1. Dual of a Dual Space
We first see that if is a vector in the vector space over the field F, then induces a linear functional on defined by
for f in .
For and ,
Hence is a linear functional.
We also see that if V is finite-dimensional and , then ; in other words, there exists a linear functional f such that . For the verification of the same, we choose an ordered basis for such that and let f be the linear functional which assigns to each vector in V its first coordinate in the ordered basis . Then, in particular, so that
Theorem 1: Let be a finite-dimensional vector space over the field . For each vector in define
, .
The mapping is then an isomorphism of V onto .
Proof
By the discussion given just before the statement of the theorem, for each the function is linear. Suppose and are in V and c is in F, and let . Then for each f in
and so .
Hence the mapping is a linear transformation from V into . Also, if and only if for all if and only if for all if and only if .
Hence the transformation is non-singular. Since
dim=dim ,
the fact that is a non-singular linear transformation from V into implies this transformation is invertible, and is therefore an isomorphism of V onto . This completes the proof of the theorem.
The following result follows immediately.
Corollary 1 Let V be a finite-dimensional vector space over the field . If L is a linear functional on the dual space of V, then there is a unique vector in V such that for every f in .
We are now ready to answer the question: whether every basis for is the dual of some basis for ?
Corollary 2 Let V be a finite-dimensional vector space over the field . Each basis for is the dual of some basis for V.
Proof: Let be a basis for . Then there is a basis for such that
.
Using Corollary 1, for each i there is a vector in V such that
for every f in , i.e., such that . It follows immediately that is a basis for V and that is the dual of this basis. This completes the proof.
In view of Theorem 1, we usually identify with and say that V ‘is’ the dual space of or that the spaces V, are naturally in duality with one another. Each is the dual space of the other.
Remark and are structurally the same. i.e., we can identify with or we can put in place of or a subspace of can be identified with corresponding subspace of .2. Annihilator of the Annihilator of a Set
We recall that if is a vector space over the field and S is a subset of , the annihilator of S is the set S0 of linear functionals f on such that for every in S.
If E is a subset of , then the annihilator is the set of linear functionals on such that for every f in E. Thus is a subset of . If we choose to identify V and as in Theorem 1, then is a subspace of V, namely, the set of all in V such that for all f in E.
We note that each subspace W is determined by its annihilator . How is it determined? The answer is that W is the subspace annihilated by all f in , that is, the intersection of the null spaces of all f in . In our present notation for annihilators, we have .
Theorem 2 If S is any subset of a finite-dimensional vector space V, then is the subspace spanned by S. i.e., if W is the subspace spanned by S, then
Proof.
Let W be the subspace spanned by S. Then clearly , hence
Therefore, what we are to prove is that . In the discussion just above the statement of the theorem, we have proved this.
Another proof of is the following:
We recall that
Also,
and since we have
so that
.
Since W is a subspace of (in the sense of isomorphism), we see that .
3. Hyperspaces
Definition 1: If V is a vector space, a hyperspace in V is a maximal proper subspace of V.
Remark If then the dimension of its hyperspace is i.e., one short of the dimension of the original space.Theorem 3: If f is a non-zero linear functional on the vector space V, then the null space of f is a hyperspace in V. Conversely, every hyperspace in V is the null space of a (not unique) non-zero linear functional on V.
Proof: Let f be a non-zero linear functional on V and its null space. Let be a vector in V which is not in , i.e., a vector such that . We will show that every vector in V is in the subspace spanned by and . That subspace consists of all vectors
in , c in F.
Let be in V. Noting that , we define
Then the vector is in since
So is in the subspace spanned by and . Hence is a hyperspace in V.
Now let N be a hyperspace in V. Fix some vector which is not in N. Since N is a maximal proper subspace, the subspace spanned by N and is the entire space V. Hence each vector in V has the form
, in N, c in F.
We now show that the vector and the scalar c are uniquely determined by . If has the representations
, in N, c in F
and
, in N , in F.
then
.
If , then which is a contradiction. Thus, and .
We conclude that, if is in V, there is a unique scalar c such that is in N. We denote that scalar by . Then it can be easily seen that g is a linear functional on V and that N is the null space of g. This completes the proof of the theorem.
Lemma If f and g are linear functionals of a vector space V, then g is a scalar multiple of f if and only if the null space of g contains the null space of f, that is, if and only if implies .
Proof: If f = 0 then g = 0 and g is trivially a scalar multiple of f. Suppose so that the null space is a hyperspace in V. Choose some vector in V with and let
.
The linear functional h = g – cf is 0 on , since both f and g are 0 there, and . Thus h is 0 on the subspace spanned by and , and that subspace is V. Hence we conclude that h = 0, i.e., that g = cf. This completes the proof of the theorem.
Theorem 4 Let g, be linear functionals on a vector space V with respective null spaces N, . Then g is a linear combination of if and only if N contains the intersection.
Proof : If and for each i, then clearly . Therefore, N contains .
Now we prove the converse by induction on the number r. By the preceding lemma, the case r = 1 is true. As induction argument, suppose the result is true for and let be linear functionals with null spaces such that is contained in N, the null space of g. Let be the restrictions of to the subspace . Then are linear functionals on the vector space . Also, if is a vector in and , then is in and so . By the induction hypothesis, there are scalars such that
Take
……………..(1)
Then h is a linear functional on and using Eq.(1), we have for every in ,
By the preceding lemma, h is a scalar multiple of . If , then
.
This completes the proof.
4. THE ADJOINT OF A LINEAR TRANSFORMATION
Suppose and W be two vector spaces over the field , and T is a linear transformation from V into W. Then we shall see that T induces a linear transformation from into .
Theorem 5 Let V and W be vector spaces over the field . For each linear transformation T from V into W, there is a unique linear transformation from into such that
for every g in and in V.
Proof:
Suppose g is a linear functional on W, and let
. . . (2)
for each in . Then (2) defines a function f from V into , namely, the composition of T, a function from into W, with g, a function from W into . Since both T and g are linear, f is linear, i.e., f is a linear functional on V. Thus T provides us with a rule which associates with each linear functional g on W a linear functional on V, defined by (2).
We now claim that is a linear transformation from into. To prove this, let and be in and c is a scalar. Then
Hence
.
This completes the proof.
We call the transpose of T or the adjoint of T.
Theorem 6 Let V and W be vector spaces over the field F, and let T be a linear transformation from V into W. The null space of is the annihilator of the range of T. If V and W are finite-dimensional, then:
(i) rank () = rank (T)
(ii) the range of is the annihilator of the null space of T.
Proof:
If g is in , then by definition
for each in V. The statement that g is in the null space of means that for every in V. Thus the null space of is precisely the annihilator of the range of T.
Suppose that V and W are finite-dimensional, say dim V = n and dim W = m. Proof of (i) : Let r be the rank of T, i.e., the dimension of the range of T. Then the annihilator of the range of T has dimension By the first statement of this theorem, the nullity of must be But then since is a linear transformation on an m-dimensional space, the rank of is and so T and have the same rank.
Proof of (ii): Let N be the null space of T. Every functional in the range of is in the annihilator of N; for, suppose for some g in ; then, if is in N ;
.
Now the range of is a subspace of the space , and
,
so that the range of must be exactly.
Theorem 7 Let V and W be finite-dimensional vector spaces over the field . Let be an ordered basis for V with dual basis, and let be an ordered basis for W with dual basis. Let T be a linear transformation from V into W; let A be the matrix of T relative to, and let be the matrix of relative to, . Then.
Proof
Let
;
.
By definition,
.
On the other hand,
For any linear functional f on V ,
.
If we apply this formula to the functional and use the fact that we obtain
,
from which it immediately follows that .
Definition 2 If A is an matrix over the field F, the transpose of A is the matrix defined by.
Theorem 7 states that if T is a linear transformation from V into W, the matrix of which in some pair of bases is A, then the transpose transformation is represented in the dual pair of bases by the transpose matrix .Theorem 8 Let A be any matrix over the field F. Then the row rank of A is equal to the column rank of A.