DLA Theory

Section 42.5 Theory

In this section.

Subsection 42.5.1 Properties of linear transformations

We begin by demonstrating that we chose the correct two axioms in Discovery 42.2 for our definition of linear transformation. We leave the verification of these properties to you, the reader.

Proposition 42.5.1. Properties of linear transformations.

A linear transformation \(\funcdef{T}{V}{W}\) satisfies the following additional linearity properties.

The image of zero is zero. That is, \(T(\zerovec_V) = \zerovec_W\text{.}\)
The image of negative is negative. That is, \(T(-\uvec{v}) = - T(\uvec{v})\) for all \(\uvec{v}\) in \(V\text{.}\)
The image of a linear combination is a linear combination. That is,
\begin{equation*} T(a_1 \uvec{v}_1 + a_2 \uvec{v}_2 + \dotsb + a_m \uvec{v}_m) = a_1 T(\uvec{v}_1) + a_2 T(\uvec{v}_2) + \dotsb + a_m T(\uvec{v}_m) \end{equation*}
for all scalars \(a_1,a_2,\dotsc,a_m\) and all vectors \(\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_m\) in \(V\text{.}\)

As is usually the case with vector spaces, a spanning set (or, preferably, a basis) tells all.

Theorem 42.5.2. Uniqueness of vector images for a domain spanning set.

Suppose \(\funcdef{T}{V}{W}\) is a linear transformation and \(\{\uvec{v}_1,\dotsc,\uvec{v}_m\}\) is a spanning set for the domain space \(V\text{.}\) Then \(T\) is uniquely determined by the image vectors \(T(\uvec{v}_1),\dotsc,T(\uvec{v}_m)\text{.}\)

In other words, \(T\) is the one unique linear transformation \(V \to W\) with these spanning set image vectors, and every other image vector for \(T\) can be determined from these spanning set image vectors.

Proof.

Write \(\uvec{w}_j\) for the \(\nth[j]\) spanning set image vector \(T(\uvec{v}_j)\text{.}\) Since each vector \(\uvec{v}\) in the domain space \(V\) can be expressed as a linear combination

\begin{equation*} \uvec{v} = a_1 \uvec{v}_1 + \dotsb + a_m \uvec{v}_m \text{,} \end{equation*}

we can use the linearity of \(T\) to compute

\begin{equation*} T(\uvec{v}) = a_1 \uvec{w}_1 + \dotsb + a_m \uvec{w}_m \text{.} \end{equation*}

This demonstrates that \(T\) is completely determined by the image vectors \(\uvec{w}_1,\dotsc,\uvec{w}_m\text{.}\)

Now assume \(\funcdef{T'}{V}{W}\) is another linear transformation with \(T'(\uvec{v}_j) = \uvec{w}_j\text{.}\) But then for

\begin{equation*} \uvec{v} = a_1 \uvec{v}_1 + \dotsb + a_m \uvec{v}_m \text{,} \end{equation*}

the linearity of \(T'\) will lead to

\begin{equation*} T'(\uvec{v}) = a_1 \uvec{w}_1 + \dotsb + a_m \uvec{w}_m = T(\uvec{v}) \text{.} \end{equation*}

Since \(T,T'\) agree on all input domain vectors, they must be the same linear transformation.

The above fact lets us easily create linear transformations from desired image vectors on a specific spanning set. (See Procedure 42.3.1.) However, as explained in Remark 42.3.2, and illustrated in Example 42.4.8, a domain space basis should be used instead of merely a spanning set.

Corollary 42.5.3. Creating a transformation from a domain space basis.

Given basis \(\basisfont{B} = \{\uvec{v}_1,\dotsc,\uvec{v}_n\}\) of vector space \(V\text{,}\) and arbitrary vectors \(\uvec{w}_1,\dotsc,\uvec{w}_n\) in vector space \(W\) (with duplicates permitted), there there exists one unique transformation \(\funcdef{T}{V}{W}\) satisfying \(T(\uvec{v}_j) = \uvec{w}_j\) for each index \(j\text{.}\)

Proof outline.

If we can show that there exists at least one such linear transformation \(\funcdef{T}{V}{W}\text{,}\) then Theorem 42.5.2 will guarantee that it is the only one.

We want \(T\) to both be linear and to satisfy \(T(\uvec{v}_j) = \uvec{w}_j\) for each index \(j\text{.}\) Given that each vector \(\uvec{v}\) in the domain space \(V\) has a unique expansion

\begin{equation*} \uvec{v} = a_1 \uvec{v}_1 + \dotsb a_n \uvec{v}_n \end{equation*}

in terms of the basis vectors in \(\basisfont{B}\) (Theorem 19.5.3), there is no ambiguity in defining \(T\) relative to such expansions:

\begin{equation*} T(\uvec{v}) = T(a_1 \uvec{v}_1 + \dotsb + a_n \uvec{v}_n) = a_1 \uvec{w}_1 + \dotsb + a_n \uvec{w}_n \text{.} \end{equation*}

Then \(T\) will be linear (check!),and since each basis vector is expanded as \(\uvec{v}_j = 1 \uvec{v}_j\text{,}\) with zero coefficients on each of the other basis vectors, we will have

\begin{equation*} T(\uvec{v}_j) = 1 \uvec{w}_j = \uvec{w}_j \text{,} \end{equation*}

as desired.

We can also use the theorem to justify our claim that every transformation \(\R^n \to \R^m\) is a matrix transformation. (And similarly for transformations \(\C^n \to \C^m\text{.}\))

Corollary 42.5.4. Transformations \(\R^n \to \R^m\) or \(\C^n \to \C^m\) are matrix transformations.

If \(\funcdef{T}{\R^n}{\R^m}\) is a linear transformation, then there exists a unique \(m \times n\) real matrix \(A\) so that \(T = T_A\text{,}\) where \(\funcdef{T_A}{\R^n}{\R^m}\) is defined by \(T_A(\uvec{x}) = A \uvec{x}\text{,}\) as usual.
If \(\funcdef{T}{\C^n}{\C^m}\) is a linear transformation, then there exists a unique \(m \times n\) complex matrix \(A\) so that \(T = T_A\text{,}\) where \(\funcdef{T_A}{\C^n}{\C^m}\) is defined by \(T_A(\uvec{x}) = A \uvec{x}\text{,}\) as usual.

Proof idea.

The proof is identical for both the real and complex versions of the statement. By Theorem 42.5.2, the transformation \(T\) is uniquely determined by the output vectors

\begin{equation*} T(\uvec{e}_1), \; T(\uvec{e}_2), \; \dotsc, \; T(\uvec{e}_n) \text{,} \end{equation*}

where \(\basisfont{S} = \{\uvec{e}_1,\dotsc,\uvec{e}_n\}\) is the standard basis for \(\R^n\) or \(\C^n\text{,}\) as appropriate. But if \(A\) is the matrix whose columns are these output vectors, then since

\begin{equation*} T_A(\uvec{e}_j) = A \uvec{e}_j \end{equation*}

is equal to the \(\nth[j]\) column of \(A\text{,}\) the transformations \(T\) and \(T_A\) agree on each of the vectors in basis \(\basisfont{S}\text{.}\) From this we can conclude that \(T\) and \(T_A\) are in fact the same transformation.

Subsection 42.5.2 Spaces of linear transformations

As in Discovery 42.7 and Subsection 42.3.6, the collection of all linear transformations from one vector space to another is itself a vector space.

Theorem 42.5.5. Transformations form a vector space.

For vector spaces \(V,W\text{,}\) collection \(L(V,W)\) of all linear transformations \(V \to W\) is a vector space under the operations

\begin{align*} (T_1 + T_2)(\uvec{v}) \amp = T_1(\uvec{v}) + T_2(\uvec{v}) \text{,} \amp (k T)(\uvec{v}) \amp = k \, T(\uvec{v})\text{.} \end{align*}

Proof.

We will verify Axiom A 1, and leave the other nine axioms to you, the reader.

If \(\funcdef{T_1,T_2}{V}{W}\) are linear, is \(\funcdef{T_1 + T_2}{V}{W}\) linear? First check additivity:

\begin{align*} (T_1 + T_2)(\uvec{v}_1 + \uvec{v}_2) \amp = T_1(\uvec{v}_1 + \uvec{v}_2) + T_2(\uvec{v}_1 + \uvec{v}_2) \amp \amp\text{(i)}\\ \amp = \bigl(T_1(\uvec{v}_1) + T_1(\uvec{v}_2)\bigr) + \bigl(T_2(\uvec{v}_1) + T_2(\uvec{v}_2)\bigr) \amp \amp\text{(ii)}\\ \amp = \bigl(T_1(\uvec{v}_1) + T_2(\uvec{v}_1)\bigr) + \bigl(T_1(\uvec{v}_2) + T_2(\uvec{v}_2)\bigr) \amp \amp\text{(iii)}\\ \amp = (T_1 + T_2)(\uvec{v}_1) + (T_1 + T_2)(\uvec{v}_2) \amp \amp\text{(iv)}\text{,} \end{align*}

with justifications

definition of addition of \(T_1,T_2\text{;}\)
additivity of \(T_1,T_2\text{;}\)
vector algebra in the codomain space \(W\text{;}\) and
definition of addition of \(T_1,T_2\text{.}\)

Now check homogeneity:

\begin{align*} (T_1 + T_2)(k \uvec{v}) \amp = T_1(k \uvec{v}) + T_2(k \uvec{v}) \amp \amp\text{(i)}\\ \amp = k \, T_1(\uvec{v}) + k \, T_2(\uvec{v}) \amp \amp\text{(ii)}\\ \amp = k \bigl(T_1(\uvec{v}) + T_2(\uvec{v})\bigr) \amp \amp\text{(iii)}\\ \amp = k \bigl((T_1 + T_2)(\uvec{v})\bigr) \amp \amp\text{(iv)}\text{,} \end{align*}

with justifications

definition of addition of \(T_1,T_2\text{;}\)
homogeneity of \(T_1,T_2\text{;}\)
vector algebra in the codomain space \(W\text{;}\) and
definition of addition of \(T_1,T_2\text{.}\)

Corollary 42.5.6. Dual spaces are vector spaces.

For every real vector space \(V\text{,}\) the dual space \(\vecdual{V} = L(V,\R)\) is a real vector space.
For every complex vector space \(V\text{,}\) the dual space \(\vecdual{V} = L(V,\C)\) is a complex vector space.

While we won't pursue a thorough study of dual spaces here, we will at least establish that a finite-dimensional vector space and its dual space have the same dimension.

Theorem 42.5.7.

Suppose \(V\) is a finite-dimensional vector space. For every basis

\begin{equation*} \basisfont{B} = \{\uvec{e}_1,\uvec{e}_2,\dotsc,\uvec{e}_n\} \end{equation*}

for \(V\text{,}\) we can define a dual basis

\begin{equation*} \vecdual{\basisfont{B}} = \{ \vecdual{\uvec{e}}_1, \vecdual{\uvec{e}}_2, \dotsc, \vecdual{\uvec{e}}_n \} \end{equation*}

for \(\vecdual{V}\) as follows.

For each index \(i\text{,}\) take \(\funcdef{\vecdual{\uvec{e}}_i}{V}{\R}\) (or \(\funcdef{\vecdual{\uvec{e}}_i}{V}{\C}\text{,}\) as appropriate) to be the unique linear transformation such that

\begin{equation*} \vecdual{\uvec{e}}_i(\uvec{e}_j) = \begin{cases} 1, \amp i = j, \\ 0, \amp i \ne j, \end{cases} \end{equation*}

for each index \(j\text{.}\)

Proof.

First, note that the existence and uniqueness of each \(\vecdual{\uvec{e}}_i\) linear functional is guaranteed by Corollary 42.5.3. So we just need to establish that these special linear functionals form a basis of \(\vecdual{V}\text{.}\)

Before we do that, it will help to establish a pattern of evaluating these special functionals. As \(\vecdual{V}\) is a vector space, a linear combination of linear functionals is also a linear functional on \(V\text{,}\) and so can be evaluated as a function on vectors in \(V\text{.}\) In particular, consider a linear combination

\begin{equation*} c_1 \vecdual{\uvec{e}}_1 + c_2 \vecdual{\uvec{e}}_2 + \dotsb + c_n \vecdual{\uvec{e}}_n \end{equation*}

evaluated at \(\uvec{e}_1\text{.}\) Using the definition of the \(\vecdual{\uvec{e}}_i\) functionals in the statement of the theorem, we calculate:

\begin{align*} \amp (c_1 \vecdual{\uvec{e}}_1 + c_2 \vecdual{\uvec{e}}_2 + \dotsb + c_n \vecdual{\uvec{e}}_n)(\uvec{e}_1) \\ \amp \phantom{(c_1} = c_1 \vecdual{\uvec{e}}_1(\uvec{e}_1) + c_2 \vecdual{\uvec{e}}_2(\uvec{e}_1) + \dotsb + c_n \vecdual{\uvec{e}}_n(\uvec{e}_1) \\ \amp \phantom{(c_1} = c_1 \cdot 1 + c_2 \cdot 0 + \dotsb + c_n \cdot 0 \\ \amp \phantom{(c_1} = c_1 \text{.} \end{align*}

Similarly,

\begin{gather} (c_1 \vecdual{\uvec{e}}_1 + c_2 \vecdual{\uvec{e}}_2 + \dotsb + c_n \vecdual{\uvec{e}}_n)(\uvec{e}_j) = c_j\label{equation-lintrans-basic-theory-lincomb-linfunc-eval}\tag{\(\star\)} \end{gather}

for each index \(j\text{.}\)

Linear independence.

To apply the Test for Linear Dependence/Independence, we begin with a homogeneous vector equation

\begin{gather} k_1 \vecdual{\uvec{e}}_1 + k_2 \vecdual{\uvec{e}}_2 + \dotsb + k_n \vecdual{\uvec{e}}_n = \zerovec\text{.}\label{equation-lintrans-basic-theory-dual-basis-lin-indep}\tag{\(\star\star\)} \end{gather}

The zero vector on the right is the zero linear functional, which evaluates to zero at all vectors in \(V\text{.}\) So evaluating both sides of (\(\star\star\)) at vector \(\uvec{e}_j\text{,}\) and using (\(\star\)) to simplify on the left, we get \(k_j = 0\) for each index \(j\text{.}\) This means that vector equation (\(\star\star\)) has only the trivial solution, so that these “dual” linear functionals are linearly independent.

Spans.

We must show that every linear functional in \(\vecdual{V}\) is a linear combination of the special \(\vecdual{\uvec{e}}_i\) functionals. So suppose that \(f\) is an arbitrary functional on \(V\text{.}\) Since \(\basisfont{B}\) is a basis for \(V\text{,}\) Theorem 42.5.2 says that \(f\) is uniquely determined by the values

\begin{equation*} f(\uvec{e}_1), \, f(\uvec{e}_2), \, \dotsc, \, f(\uvec{e}_n) \text{.} \end{equation*}

Let \(a_1,a_2,\dotsc,a_n\) represent these values, and let \(g\) be the linear functional

\begin{equation*} a_1 \vecdual{\uvec{e}}_1 + a_2 \vecdual{\uvec{e}}_2 + \dotsb + a_n \vecdual{\uvec{e}}_n \text{.} \end{equation*}

The pattern of (\(\star\)) says that

\begin{equation*} g(\uvec{e}_j) = a_j \text{.} \end{equation*}

But since \(f\) is uniquely determined by the \(a_j\) values, we can conclude that \(f = g\text{,}\) a linear combination of the \(\vecdual{\uvec{e}}_i\) functionals.

Corollary 42.5.8.

For finite-dimensional \(V\text{,}\) we have \(\dim \vecdual{V} = \dim V\text{.}\)