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Section 43.5 Theory

Subsection 43.5.1 Kernel and image are subspaces

Apply the Subspace Test.

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

\begin{equation*} T(\uvec{v}_1 + \uvec{v}_2) = T(\uvec{v}_1) + T(\uvec{v}_2) = \zerovec_W + \zerovec_W = \zerovec_W\text{,} \end{equation*}

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

\begin{equation*} T(k \uvec{v}) = k T(\uvec{v}) = k \zerovec_W = \zerovec_W \text{,} \end{equation*}

so yes, \(k \uvec{v}\) is also in \(\ker T\text{.}\)

Apply the Subspace Test.

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

\begin{align*} T(\uvec{v}_1) \amp = \uvec{w}_1 \text{,} \amp T(\uvec{v}_2) \amp = \uvec{w}_2\text{.} \end{align*}

Is \(\uvec{w}_1 + \uvec{w}_2\) also in \(\im T\text{?}\) Using the additivity of \(T\text{,}\) we have

\begin{equation*} \uvec{w}_1 + \uvec{w}_2 = T(\uvec{v}_1) + T(\uvec{v}_2) = T(\uvec{v}_1 + \uvec{v}_2)\text{,} \end{equation*}

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

\begin{equation*} T(\uvec{v}) = \uvec{w} \text{.} \end{equation*}

Is \(k \uvec{w}\) also in \(\im T\) for arbitrary scalar \(k\text{?}\) Using the homogeneity of \(T\text{,}\) we have

\begin{equation*} k \uvec{w} = k T(\uvec{v}) = T(k \uvec{v}) \text{,} \end{equation*}

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.

Now we will justify the conclusion of Procedure 43.3.1.

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

\begin{equation*} c_1 T(\uvec{v}_1) + c_2 T(\uvec{v}_2) + \dotsb + c_r T(\uvec{v}_r) = \zerovec_W \text{.} \end{equation*}

The linearity of \(T\) can be used to collapse the linear combination on the left into a single image vector

\begin{equation*} T(c_1 \uvec{v}_1 + c_2 \uvec{v}_2 + \dotsb + c_r \uvec{v}_r) = \zerovec_W \text{,} \end{equation*}

which implies that the domain space linear combination

\begin{equation*} c_1 \uvec{v}_1 + c_2 \uvec{v}_2 + \dotsb + c_r \uvec{v}_r \end{equation*}

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

\begin{equation*} c_1 \uvec{v}_1 + c_2 \uvec{v}_2 + \dotsb + c_r \uvec{v}_r = \zerovec_V \text{.} \end{equation*}

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:

\begin{equation*} c_1 = c_2 = \dotsb = c_r = 0 \text{.} \end{equation*}

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

\begin{equation*} T(\basisfont{B}_V) = \{ T(\uvec{u}_1), \dotsc, T(\uvec{u}_k), T(\uvec{v}_1), \dotsc, T(\uvec{v}_r) \} \end{equation*}

is a spanning set for \(\im T\text{.}\) But since the \(\uvec{u}_j\) are in \(\ker T\text{,}\) we actually have

\begin{equation*} T(\basisfont{B}_V) = \{ \zerovec_W, \dotsc, \zerovec_W, T(\uvec{v}_1), \dotsc, T(\uvec{v}_r) \}\text{,} \end{equation*}

and clearly the collection of image vectors

\begin{equation*} T(\basisfont{K}') = \{ T(\uvec{v}_1), \dotsc, T(\uvec{v}_r) \} \end{equation*}

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.

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

\begin{gather*} \dim (\ker T) = \# \basisfont{K} \text{,} \\ \dim (\im T) = \# T(\basisfont{K}') = \# \basisfont{K}' \text{,} \\ \dim V = \# \basisfont{K} + \# \basisfont{K}' \text{,} \end{gather*}

and the result follows.