Section 43.5 Theory
In this section.
Subsection 43.5.1 Kernel and image are subspaces
Theorem 43.5.1.
For linear transformation \(\funcdef{T}{V}{W}\text{,}\) \(\ker T\) is a subspace of the domain space \(V\) and \(\im T\) is a subspace of the codomain space \(W\text{.}\)
Proof: Kernel is a subspace.
Apply the Subspace Test.
Nonempty.
By Statement 1 of Proposition 42.5.1, \(\ker T\) always contains \(\zerovec_V\text{,}\) the zero vector in the domain space \(V\text{.}\)
Closed under addition.
Suppose \(\uvec{v}_1,\uvec{v}_2\) are in \(\ker T\text{.}\) Is \(\uvec{v}_1 + \uvec{v}_2\) also in \(\ker T\text{?}\) Using the additivity of \(T\text{,}\) we have
so yes, \(\uvec{v}_1 + \uvec{v}_2\) is also in \(\ker T\text{.}\)
Closed under scalar multiplication.
Suppose \(\uvec{v}\) is in \(\ker T\text{.}\) Is \(k \uvec{v}\) also in \(\ker T\) for arbitrary scalar \(k\text{?}\) Using the homogeneity of \(T\text{,}\) we have
so yes, \(k \uvec{v}\) is also in \(\ker T\text{.}\)
Proof: Image is a subspace.
Apply the Subspace Test.
Nonempty.
By Statement 1 of Proposition 42.5.1, \(\im T\) always contains \(\zerovec_W\text{,}\) the zero vector in the codomain space \(W\text{.}\)
Closed under addition.
Suppose \(\uvec{w}_1,\uvec{w}_2\) are in \(\im T\text{,}\) so that there exist \(\uvec{v}_1,\uvec{v}_2\) in \(V\) with
Is \(\uvec{w}_1 + \uvec{w}_2\) also in \(\im T\text{?}\) Using the additivity of \(T\text{,}\) we have
so yes, \(\uvec{w}_1 + \uvec{w}_2\) is also in \(\im T\text{.}\)
Closed under scalar multiplication.
Suppose \(\uvec{w}\) is in \(\im T\text{,}\) so that there exists \(\uvec{v}\) in \(V\) with
Is \(k \uvec{w}\) also in \(\im T\) for arbitrary scalar \(k\text{?}\) Using the homogeneity of \(T\text{,}\) we have
so yes, \(k \uvec{w}\) is also in \(\im T\text{.}\)
Subsection 43.5.2 Basis and dimension of kernel and image
First we formally state the pattern of Discovery 43.5, which we discussed further in Subsection 43.3.2.
Lemma 43.5.2. Image of a spanning set is a spanning set for the image.
If \(\funcdef{T}{V}{W}\) is linear and
is a spanning set for the domain space \(V\text{,}\) then the collection of image vectors
is a spanning set for \(\im T\text{.}\)
Now we will justify the conclusion of Procedure 43.3.1.
Theorem 43.5.3. Basis for image.
Suppose \(\funcdef{T}{V}{W}\) is linear with \(V\) finite-dimensional, and
are collections of vectors in \(V\) so that
- \(\basisfont{K}\) is a basis for \(\ker T\text{,}\) and
- the vectors of \(\basisfont{K},\basisfont{K}'\text{,}\) taken all together, form a basis for \(V\text{.}\)
Then the collection of image vectors
is a basis for \(\im T\text{.}\)
Proof.
We need to establish that the collection \(T(\basisfont{K}')\) is both linearly independent and a spanning set for \(\im T\text{.}\)
Linear independence.
To apply the Test for Linear Dependence/Independence, we begin with a homogeneous vector equation
The linearity of \(T\) can be used to collapse the linear combination on the left into a single image vector
which implies that the domain space linear combination
is in \(\ker T\text{.}\) As \(\ker T = \Span \basisfont{K}\text{,}\) this means that the above linear combination is simultaneously in both of the subspaces \(\Span \basisfont{K}\) and \(\Span \basisfont{K}'\text{.}\) However, these spaces are independent, since we have assumed that their bases (\(\basisfont{K}\) and \(\basisfont{K}'\)) together form a basis for \(V\text{,}\) hence these two bases form an independent set when taken together. By Theorem 28.6.4, a pair of independent spaces can only intersect at the zero vector, so we may conclude that
As the vectors in \(\basisfont{K}'\) are assumed to be independent (since they form part of a basis for \(V\)), the only way this last vector equation above is possible is the trivial way:
In particular, since all of the \(c_j\) scalars must be zero, we can conclude that the vectors in \(T(\basisfont{K}')\) are independent, as desired.
Spans.
Write \(\basisfont{B}_V\) for the collection of all vectors from \(\basisfont{K}\) and \(\basisfont{K}'\) taken together, which we have assumed is a basis for the domain space \(V\text{.}\) By Lemma 43.5.2, the collection of image vectors
is a spanning set for \(\im T\text{.}\) But since the \(\uvec{u}_j\) are in \(\ker T\text{,}\) we actually have
and clearly the collection of image vectors
will span the same space that \(T(\basisfont{B}_V)\) does.
Finally, we can connect the dimensions of the kernel and image of a transformation.
Theorem 43.5.4. Dimension Theorem.
Suppose \(\funcdef{T}{V}{W}\) is linear with \(V\) finite-dimensional. Then the sum of the rank and nullity of \(T\) is the dimension of the domain space \(V\text{.}\) That is,
Proof.
Let \(\basisfont{K}\) be a basis for \(\ker T\text{,}\) and let \(\basisfont{K}'\) be a collection of vectors in \(V\) that enlarges \(\basisfont{K}\) to a basis \(\basisfont{B}_V\) of \(V\) (Proposition 20.5.4). Then Theorem 43.5.3 says that the collection of image vectors \(T(\basisfont{K}')\) is a basis for \(\im T\text{.}\) Now, the collection \(T(\basisfont{K}')\) cannot contain any duplicates because it is linearly independent. So \(T(\basisfont{K}')\) contains exactly the same number of vectors as \(\basisfont{K}'\text{.}\)
Using the notation \(\# S\) to mean the number of vectors in collection \(S\text{,}\) we now have
and the result follows.