Page 126 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes
then T c 1 1 c 2 2 ... c k k 0
and since T is non-singular

c 1 1 c 2 2 ... c k k 0

from which it follows that each c i 0, because S is an independent set. The argument shows that
the image of S under T is independent.

Suppose that T carries independent subsets onto independent subsets. Let be a non-zero
vector in V. Then the set S consisting of the one vector is independent. The image of S is the set
consisting of the one vector T , and this set is independent. Therefore T 0, because the set
consisting of the zero vector alone is dependent. This shows that the null space of T is zero
subspace i.e., T is non-singular.
Thus in deciding whether S is independent it does not matter whether we look at S or T(S). From
this one sees that an isomorphism is ‘dimension preserving’, that is any finite-dimensional
subspace of V has the same dimension as the image under T. Here is a very simple illustration
of this idea. Suppose A is an m×n matrix over the field F. We have really given two definitions
n
,
of the solution space of the matrix A. The first is the set of all n-tuples x x 2 ...x n in F which
1
satisfy each of the equations in the system AX = 0. The second is the set of all n × 1 column
matrices X such that AX = 0. The first solution space is thus a subspace of F and the second is a
n
subspace of the space of all n×1 matrices over F. Now there is a completely obvious isomorphism
n
between F and F , namely
n+1
x
1
x
x 1 ,x 2 ,...x n  2 .
x n

Under this isomorphism, the first space of A is carried onto the second solution space. These
spaces have the same dimension, and so if we want to prove a theorem about the dimension of
the solution space, it is immaterial which space we choose to discuss.

(n)
(m)
Example 1: F is isomorphic F if and only if n = m.
(n)
Proof: Here F has, as one basis, the set of n vectors (1, 0, 0, …, 0), (0, 1, …, 0), … (0, 0, …, 1).
Likewise F has a basis containing m vectors. An isomorphism maps a basis of F onto a basis
(n)
(m)
(m)
(n)
(m)
of F . This is only possible if the dimensions of F and F are the same. Hence n = m.
Example 2: Prove that
(n)
(1)
(a) F is not isomorphic to F for n > 1.
(b) F is not isomorphic to F .
(3)
(2)
Example 3: Let V = C be the set of complex numbers, remembering only the addition of
two elements as + and multiplication r of a complex element by a real number. Then the
linear transformation T mapping R C sending (a, b) a + b i is an isomorphism.
2

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