Page 120 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Proof: (a) This property of the identity function is obvious. We have stated here merely for
emphasis.
(b)

[U(T + T ) ]( ) = U[(T + T ) ( )]
1 2 1 2
= U(T + T )
1 2
= U(T )] + U (T )
1 2
= (UT )( ) + (UT )( )
1 2
so that
U(T + T ) = UT + UT
1 2 1 2
Also [(T + T ) U]( ) = (T + T ) (U )
1 2 1 2
= T (U )] + T (U )
1 2
= (T U)( ) + (T U)( )
1 2
so that (T + T ) U = T U + T U.
1 2 1 2
(c) It is easy to prove (c) in a simple way.
Non-singular Transformations

A linear transformation T from v and w is said to be non-singular transformation if and only if
there exists a mapping T* from R onto v such that TT* = I, where I is the identity mapping on V.
T
–1
–1
–1
–1
Thus T* = T . Thus TT = T T = I, T is called inverse of T.
The function T from v into w is called invertible if there exists a function U from w into v such that
UT is the identity function on v and TU is the identity function on w. If T is invertible, the
function U is unique and is denoted by T . Further more T is invertible if and only if
–1
1. T is 1:1, that is, T = T implies = ;
2. T is onto, that is, the range of T is w.
Theorem 7: Let v and w be vector spaces over the field F and let T be a linear transformation from
v into w. If T is invertible, then the inverse function T is a linear transformation from w onto v.
–1
Proof: What we are proving here is that if a linear transformation T is invertible, then the
inverse T is also linear.
–1
Let and be vectors in w and let c be a scalar. We wish to show that
1 2
–1
–1
T (C + ) = CT –1 + T
1 2 1 2
Let = T –1 , i = 1, 2, that is, let be the unique vector in v such that T = . Since T is linear,

i i i i i
T(C + ) = CT + T
1 2 1 2
= C + .
1 2
Thus C + is the unique vector in v which is sent by T into C + and so
1 2 1 2
T (C + ) = C +
–1
1 2 1 2
= CT –1 + T –1
1 2
and thus T is linear.
–1

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