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Section 43.3 Concepts

Subsection 43.3.1 Kernel and image of matrix transformations

As we identified in Discovery 43.1 and Discovery 43.4, we already know what kernel and image mean for a matrix transformation \(\funcdef{T_A}{\R^n}{\R^m}\text{,}\) because we have studied them before, albeit with different terminology.

A vector \(\uvec{x}\) in \(\R^n\) is in \(\ker T_A\) precisely when \(A \uvec{x} = \zerovec\text{,}\) which is the same as saying that \(\uvec{x}\) is in the null space of \(A\text{.}\) And a vector \(\uvec{b}\) in \(\R^m\) is in \(\im T_A\) precisely when there exists at least one corresponding solution \(\uvec{x}\) to the matrix equation \(A \uvec{x} = \uvec{b}\text{.}\) Since a matrix-times-vector product can be considered as a linear combination of the columns of \(A\) (see Subsection 22.3.2), we can say that \(\uvec{b}\) is in \(\im T\) precisely when it is in the column space of \(A\text{.}\)

Also in Discovery 43.1 and Discovery 43.4, we reminded ourselves that the dimension of the column space and the dimension of the null space always add up to the number of columns in the matrix. This is because the dimension of the column space is the number of leading ones in the RREF of the matrix, while the dimension of the null space is the number of parameters required to solve the homogeneous system, where every variable that does not have a leading one in its column will be assigned a parameter. But we already knew this as the Rank-Nullity Theorem. As matrix transformations are our model, it shouldn't be too surprising that we shall soon see a similar pattern for the dimensions of the kernel, image, and domain space of a general linear transformations.

Subsection 43.3.2 Bases for kernel and image

For a matrix transformation, we already have procedures for determining bases for the null space and the column space of a linear transformation. (See Chapter 21.) Determining a basis for a kernel will be very similar, but for an image we will need a new tactic.

Basis for kernel.

The kernel of a linear transformation \(\funcdef{T}{V}{W}\) is defined by a homogeneous condition:

\begin{gather} T(\uvec{x}) = \zerovec_W\text{.}\label{equation-lintrans-ker-im-concepts-homog-ker-condition}\tag{\(\star\)} \end{gather}

Unravelling this condition in terms of the type of domain vector \(\uvec{x}\text{,}\) the definition of the transformation \(T\text{,}\) and the type of zero vector \(\zerovec_W\) in the codomain space should lead to a homogeneous linear system to solve. Similarly to determining a basis for a null space, solving the homogeneous system that arises from (\(\star\)) can be carried out by simplifying the system (probably via row reducing some matrix), assigning parameters, and then extracting the basis vector attached to each parameter. See the portions of Example 43.4.2 and Example 43.4.3 where a parametric description of kernel vectors was used to determine a basis for a transformation kernel.

Spanning set for image.

As we explored in Discovery 43.5, a spanning set for the domain space of a linear transformation can be pushed forward to a spanning set for the image. That is, if \(\funcdef{T}{V}{W}\) is linear, and \(V = \Span S\text{,}\) then \(\im T\) is spanned by \(T(S)\text{,}\) the collection of image vectors \(T(\uvec{v})\) for all \(S\)-vectors \(\uvec{v}\text{.}\) This is because if

\begin{equation*} \uvec{u} = a_1 \uvec{v}_1 + \dotsb + a_\ell \uvec{v}_\ell \end{equation*}

is an expansion of vector \(\uvec{u}\) in \(V\) in terms of some collection of \(S\)-vectors \(\uvec{v}_1,\dotsc,\uvec{v}_\ell\text{,}\) then the linearity of \(T\) gives us

\begin{equation*} T(\uvec{u}) = a_1 T(\uvec{v}_1) + \dotsb + a_\ell T(\uvec{v}_\ell) \text{,} \end{equation*}

a linear combination of the image vectors in \(T(S)\text{.}\) However, even if \(S\) is a linearly independent spanning set (and so a basis for \(V\)), there is no guarantee that the image vectors \(T(\uvec{v})\) in \(T(S)\) will remain independent. So to get a basis for \(\im T\) in this way, we will need to choose a spanning set \(S\) for the domain \(V\) in a more deliberate way.

Basis for image.

Suppose we have linear \(\funcdef{T}{V}{W}\text{,}\) and we know \(V = \Span S\) for

\begin{equation*} S = \{ \uvec{v}_1, \uvec{v}_2, \dotsc, \uvec{v}_\ell \} \text{.} \end{equation*}

Then as in the discussion above, we will also have

\begin{equation*} \im T = \Span \{ T(\uvec{v}_1), T(\uvec{v}_2), \dotsc, T(\uvec{v}_\ell) \} \text{.} \end{equation*}

But every finite spanning set can be reduced to a basis (Proposition 19.5.1), so there should be a way to reduce the collection of image vectors

\begin{equation*} T(S) = \{ T(\uvec{v}_1), T(\uvec{v}_2), \dotsc, T(\uvec{v}_\ell) \} \end{equation*}

to a basis for \(\im T\text{.}\) In particular examples we could do this through calculation via the Test for Linear Dependence/Independence. But we can make the task of determining dependent vectors to discard from \(T(S)\) much simpler if we choose the original spanning set \(S\) more deliberately.

In particular, suppose we have a basis

\begin{equation*} \basisfont{K} = \{ \uvec{u}_1, \dotsc, \uvec{u}_k \} \end{equation*}

for \(\ker T\) in \(V\text{.}\) We can enlarge this to a basis

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

for \(V\) (Proposition 20.5.4). Then, as in previous discussion above,

\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*}

will be a spanning set for \(\im T\text{.}\) But since the \(\uvec{u}_j\) came from the kernel, we really have

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

and it is obvious that we can reduce this spanning set by discarding all those zero vectors:

\begin{equation*} S' = \{ T(\uvec{v}_1), \dotsc, T(\uvec{v}_r) \} \text{.} \end{equation*}

We will still have \(\im T = \Span S'\text{,}\) and in fact we will see that building \(\basisfont{B}_V\) to have as many linearly independent vectors from \(\ker T\) as possible will ensure that \(S'\) is a linearly independent spanning set (hence a basis) for \(\im T\text{.}\)

Again, see the examples in Subsection 43.4.2 where the above procedure is carried out.

Subsection 43.3.3 Dimensions of the kernel and image of a transformation

Procedure 43.3.1 tells us about the relationship between the dimensions of \(\ker T\) and \(\im T\) for linear \(\funcdef{T}{V}{W}\text{,}\) with \(V\) finite-dimensional. Using the notation in the procedure, the vectors in \(\basisfont{K}\) and \(\basisfont{K}'\) together create a basis for \(V\text{,}\) so the numbers of vectors in these collections add up to the dimension of \(V\text{.}\) Collection \(\basisfont{K}\) is a basis for \(\ker T\text{,}\) so the number of vectors in \(\basisfont{K}\) is the dimension of \(\ker T\text{.}\) The collection of image vectors \(T(\basisfont{K}')\) is a basis for \(\im T\text{,}\) and has the same number of vectors \(\basisfont{K}'\text{.}\) Putting these two facts together, we have the linear transformation version of Rank-Nullity Theorem: the nullity of \(T\) and the rank of \(T\) add to the dimension of the domain space \(V\).