Introduction
As it was shown in previous sections discussed in class, the set Mmn of all m×n matrices is a vector space under the operations of matrix scalar multiplication and addition. In this paper, we will describe the set of all linear transformations of an n-dimensional vector space V into an m-dimensional vector space W, which is also a vector space U. This is the vector space under two defined operations. The relationship between Mmn and U will be also described.
Body
Consider V and W – two vector spaces of the dimensions n and m, respectively. Let the following linear transformations are defined:
L1:V→WL2:V→W
Define a mapping L:V→W as follows:
L1:V→WLx=L1x+L2x, ∀x∈VDenote L=L1⊞L2 – the sum of L1 and L2. Let also L3:V→W is a linear transformation for the mapping H:V→W defined as Hx=cL3x, c∈R. Denote H as c⊡L3 and call it the scalar multiple of L3 by c.
Example #1
Consider the following example. Let V=R3, W=R2, L1:R3→R2, L2:R3→R2 defined as:
L1x=L1u1, u2, u3=u1+u2, u2+u3L2x=L2u1, u2, u3=u1+u3, u2
Hence
L1⊞L2x=2u1+u2+u3, 2u2+u3
And
3⊡L1x=3u1+3u2, 3u2+3u3
This example shows how scalar multiple and the sum of the linear transformations work.
Example #2
Consider the following linear transformations:
L1:R2→R3, L2:R2→R3, L3:R2→R3L1u1, u2=u1+u2, 2u2+u2L2u1,u2=u2-u1, 2u1+u2+u1L3u1, u2=3u1-2u2, u1+2u2
The task is to determine if S={L1,L2,L3} is linearly independent.
Solution:
Consider that
a1⊡L1⊞a2⊡L2⊞a3⊡L3=O, a1, a2,a3∈R
Then for e1T=[1,0]
a1⊡L1⊞a2⊡L2⊞a3⊡L3e1T=Oe1T=0,0,0
Hence
a1L1e1T+a2L2e2T+a3L3e1T=a11,2,0+a2-1,2,1+a33,0,1=0,0,0
The problem is now to solve the homogeneous system:
1-13220011a1a2a3=000
Obviously, the solution is one and only a1=a2=a3=0.Hence, S is linearly independent
Theorem #1
Consider U – the vector space of all linear transformations of an n-dimensional vector space V into an m-dimensional vector space W, n≠0, m≠0 under the operations ⊞ and ⊡. Then U is isomorphic to the vector space Mmn of all m×n matrices.
Proof
Let S={v1, v2,,vn} and T={w1,w2,,wm} be ordere bases for V and W, respectively. Define a function M:U→Mmn by letting M(L) be the matrix representing L with respect to the bases S and T. Let’s show that M is an isomorphism.
Obviously, M is one-to-one. If L1 and L2 are two different elements of U, then L1vj≠L2vj for some J=1,2,,n. This means that the jth columns of ML1 and M(L2), which are the coordinate vectors of L1(vj) and L2(vj), respectively, with respect to T, are different, so M(L1)≠M(L2). Thus, M is one-to-one.
M is also onto. Let A=[aij] - a given m×n matrix. Hence, A is an element of Mmn. Define a function:
L:V→W
Lvi=k=1makiwk, i=1,2,n
And if x=c1v1+c2v2++cnvn, define L(x) by:
Lx=k=1nciLvi
Obviously, L is a linear transformation and the matrix representing L with respect to S and T is A=[aij]. Thus, ML=A, so M is onto.
Let ML1=A=[aij] and ML2=B=[bij]. We show that ML1⊞L2=A+B. Note that the jth column of M(L1⊞L2) is:
L1⊞L2vjT=L1vj+L2vjT=L1vjT+L2vjT
Thus, the jth column of M(L1⊞L2) is the sum of the jth columns of ML1=A and ML2=B. Hence, ML1⊞L2=A+B.
Finaly, let ML=A and c be a real number. According to the ideas given in the previous paragraph, Mc⊡L=cA. Hence, U and Mmn are isomorphic. Q.E.D.
As a consequence, the dimension of U is m*n, for dimMmn=mn. It means that when dealing with finite-dimensional vector spaces, we can replace all linear transformations by their matrices and work with them in matrix form. Also, it should be noted that matrices lend themselves much more readily than linear transformations to computer implementations.
Example #3
Consider matrix A:
A=12-12-13,
Also, let S and T are natural bases for R3 and R2, respectively:
S=e1, e2, e3T={e1, e2}
Find the unique linear transformation L:R3→R2, whose representation with respect to S and T is A.
Let S’ and T’ are ordered bases for R3 and R2, respectively:
S'=101,110, 011T'=13, 2-1
Determine the linear transformation L:R3→R2 whose representation with respect to S’ and T’ is A.
Compute L123, using L as determined in part (b).
Solution
a)
Le1=1e1+2e2=12,Le2=2e1-1e2=2-1,Le3=-e1+3e2=-13,
If x=[a1a2a3] is in R3, then define L(x) by:
Lx=La1e1+a2e2+a3e3=a1Le1+a2Le2+a3Le3Lx=a112+a22-1+a3-13=a1+2a2-a32a1-a2+3a3
Consider that
Lx=12-12-13a1a2a3
So we could have defined L by Lx=Ax, x∈R3. It is possible when S and T are the natural bases.
b)
L101=113+22-1=51
L110=213-12-1=07
L011=-113+32-1=5-6
Then if x=a1a2a3, we express x in terms of the basis S’ as:
x=b1101+b2110+b3011, xS'=b1b2b3
Define L(x) by:
Lx=b1L101+b2L110+b3L011=b151+b207+b35-6==5b1+5b3b1+7b2-6b3
c)
In order to find L123, we can expand this vector as:
123=1101+0110+2011
Thus
123S'=102
Substitute these values in 5b1+5b3b1+7b2-6b3:
5*1+5*21+7*0-6*2=15-11
Note that the linear transformations obtained in this example depend on the ordered basis for R2 and R3.
Consider V1be an n-dimensional vector space, V2 is an m-dimensional vector space and V3 is a p-dimensional vector space. Let L1:V1→V2, L2:V2→V3 are linear transformations. Define the composite function:
L2∘L1:V1→V3 by L2∘L1x=L2L1x, x∈V1.
This function is also a linear transformation.
Theorem #2
Let V1 be an n-dimensional vector space, V2 is an m-dimensional vector space and V3 is a p-dimensional vector space. Let L1:V1→V2, L2:V2→V3 are linear transformations. If the ordered bases P, S and T are chosen for V1, V2 and V3, respectively, then ML2∘L1=ML2M(L1)
