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Lecture 16

# Lecture 16.pdf

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University of Waterloo

Mathematics

MATH 136

Robert Sproule

Winter

Description

Monday, February 10 − Lecture 16 : Subspaces associated to linear mappings
Concepts:
1. Kernel of a linear mapping.
2. Range of a linear mapping
3. Codomain of a linear mapping
4. Kernel of a linear mapping viewed as a subspace.
16.1 Definition − Let T: ℝ → ℝ be a linear mapping. The range, Range(T), of T is a
m
subset of ℝ defined as follows:
m n
Range(T) = {y ∈ ℝ : y = T(x) for some x in ℝ }.
The set is referred to as the codomain of T. The notion of the “range” does not only apply
to linear mappings. It is used in relation to any function. Every function f : A → B has a
range which is a subset of Bdefined precisely as stated above.The range of a function
can be a proper subset of the codomain.
16.1.1 Observation – For any linear mapping T, 0 belongs to the range of T.
Notice that if T : ℝn→ ℝ is a linear mapping then 0-vector in ℝ must belong to the
n m
Range(T) since if u ∈ ℝ then T(0u) = 0T(u) = 0 ∈ ℝ . So, for any linear mapping T,
0 belongs to the range of T.
n m
16.1.2 Theorem – If T : ℝ → ℝ is a linear mapping the range(T) is a subspace of
ℝ .
Proof : Since Range(T) contains the 0-vector it is non-empty. Let u and v be vectors in
range(T). Then there exists a and bin such that T(a) = u and T(b) = v. Since T is
linear, αu + βv = αT(a) + βT(b) = T(αa + βb). So αu + βv ∈ Range(T). So Range(T)
is closed under linear combinations and so is a subspace.
16.1.3 Theorem – Let A m×n = [c 1 …2c ] n m×n be the matrix induced by the linear
mapping T : ℝ → ℝ . Then Range(T) = Span{c c … c }, a subspace of ℝ . m
1 2 n
Proof :
- Given: That T(x) = Ax = [c c … c ]x for all x in ℝ . n
1 2 n
- The range of T is the set all vectors y in ℝ such T(x) = y for some x in ℝ . Then
m
the range of T is the set of y in ℝ such that Ax = y has a solution.
- The matrix equation Ax = d has a solution if and only if Ab = d for

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