Linear Algebra: UNIT II: Linear Transformations and Diagonalization
Null space (or) Kernel N(T) and Range (or) Image R(T): Theorem Part 2
Important Theorems for Engineering Maths or Mathematics - Null space (or) Kernel N(T) and Range (or) Image R(T): Theorem Part 2
Theorem 6
Let V and W be vector spaces
over F and suppose that {v1, v2, … vn} is a
basis for V. For w1, W2,..., wn in W, there
exists exactly one linear transformation T:V→W such that T (vi) = wi
for i = 1, 2, ... n.
Proof:
Let
x ∈ V, then
x=
aivi, where
ai ∈
F and also
a1, a2, …., an
are unique scalars.
Define
T: V→W by T(x) = aiwi.
(a)
To prove: T is linear,
Suppose
that u, v ∈
V and d ∈ F, then we may write
u
= bivi and
v
= civi
for
some scalars b1, b2,… bn,
c1, c2, ... cn
Thus
du+v = dbivi + civi
= (dbi + ci)vi
T (du+v) = T ( (dbi +ci)
vi
= (dbi +ci) T (vi)
= (dbi +ci) wi
= dbiwi + ciwi
=
dT(u) + T(v)
T
is linear.
(b) Clearly T(vi) = wi
for i = 1, 2, ..... n.
(c)
To prove: T is unique
Suppose
that U: V→ W is linear and U (vi) = wi
for
i = 1, 2, ... n. Then for x ∈
V with
x= aίvi
we
have
U
(x) = aiU(vi) = aiwi
=
T(x)
Hence
U = T
T is unique.
Linear Algebra: UNIT II: Linear Transformations and Diagonalization : Tag: maths, mathematics : - Null space (or) Kernel N(T) and Range (or) Image R(T): Theorem Part 2
Linear Algebra: UNIT II: Linear Transformations and Diagonalization
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Linear Algebra
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