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Section 45.5 Theory

Subsection 45.5.1 The matrix of a linear transformation

Once a choice of bases for domain and codomain are made, there is one single matrix that will represent a transformation.

We already discussed this in Subsection 45.3.1. The matrix of \(T\) relative to \(\basisfont{B},\basisfont{B}'\) is defined to be the standard matrix of the composition \(\coordmap{B'} T \invcoordmap{B}\text{,}\) where \(\coordmap{B},\coordmap{B'}\) are the coordinate maps on \(V,W\) relative to \(\basisfont{B},\basisfont{B}'\text{,}\) respectively. And we know that standard matrices are unique from Corollary 42.5.4.

If the bases for domain and codomain are made in an informed manner, the matrix for the transformation will be particularly simple.

We already discussed this in Subsection 45.3.3. The matrix \(\matrixOf{T}{B'B}\) will have this form when we choose \(\basisfont{B}\) to be an extension of a basis for \(\ker T\) to a basis for \(V\) (with the kernel basis vectors appearing after all the non-kernel vectors), and we choose \(\basisfont{B}'\) to be an extension of the basis for \(\im T\) afforded by the nonzero image vectors in \(T(\basisfont{B})\) to a basis for \(W\) (with the image basis vectors appearing before all the non-image vectors).

An identity operator does not transform the vectors in the domain space, but its matrix can change the basis relative to which coordinate vectors are formed. As this was already explored in Discovery 45.6 and examined in detail in Subsection 45.3.3, we state it without proof

Subsection 45.5.2 Properties of a transformation from properties of its matrix

  1. By definition, vector \(\uvec{v}\) in \(V\) is in \(\ker T\) precisely when \(T(\uvec{v}) = \zerovec_W\text{.}\) As the coordinate map is an isomorphism, the only vector in \(W\) with coordinate vector \(\zerovec_m\) (where \(m = \dim W\)) is \(\zerovec_W\text{.}\) So \(T(\uvec{v}) = \zerovec_W\) if and only if
    \begin{equation*} \matrixOf{T(\uvec{v})}{B'} = \zerovec_m \text{.} \end{equation*}
    Using (\(\star\)), we can say that \(\uvec{v}\) is in \(\ker T\) if and only if
    \begin{equation*} \matrixOf{T}{B'B} \matrixOf{\uvec{v}}{B} = \zerovec_m \text{,} \end{equation*}
    which says that \(\matrixOf{\uvec{v}}{B}\) is in the null space of \(\matrixOf{T}{B'B}\text{.}\)
  2. It follows from Statement 1 that the isomorphism \(\coordmap{B}\) must send a basis for \(\ker T\) to a basis for the null space of \(\matrixOf{T}{B'B}\text{.}\) Hence the dimensions of these two spaces must be equal.
  3. By definition, vector \(\uvec{w}\) in \(W\) is in \(\im T\) precisely when there exists at least one vector \(\uvec{v}\) in \(V\) with \(\uvec{w} = T(\uvec{v})\text{.}\) In this case, using (\(\star\)) and the fact that \(\coordmap{B'}\) is an isomorphism, we would have
    \begin{equation*} \matrixOf{\uvec{w}}{B'} = \matrixOf{T(\uvec{v})}{B'} = \matrixOf{T}{B'B} \matrixOf{\uvec{v}}{B} \text{.} \end{equation*}
    But then, using the fact that \(\coordmap{B}\) is also an isomorphism, we can say that \(\uvec{w}\) is in \(\im T\) if and only if there exists a column vector \(\uvec{x}\) with
    \begin{equation*} \matrixOf{\uvec{w}}{B'} = \matrixOf{T}{B'B} \uvec{x} \text{.} \end{equation*}
    As the column space of a matrix \(A\) consists of the results of all possible products \(A \uvec{x}\) (Subsection 21.3.1), we arrive at the statement at hand.
  4. Again, it follows from Statement 3 that the isomorphism \(\coordmap{B'}\) must send a basis for \(\im T\) to a basis for the column space of \(\matrixOf{T}{B'B}\text{.}\) Hence the dimensions of these two spaces must be equal.

We can also characterize isomorphisms via invertibility of their matrices.

Statement 1 implies Statement 2.

Assume \(T\) is an isomorphism, and suppose \(\basisfont{B},\basisfont{B}'\) is an arbitrary choice of bases for \(V,W\text{,}\) respectively. Then the dimensions of \(V\) and \(W\) must be equal (Corollary 44.5.15), hence \(\matrixOf{T}{B'B}\) must be square. Furthermore, to be an isomorphism the transformation \(T\) must be injective, hence its kernel is trivial (Theorem 44.5.5). Using Statement 1 of Proposition 45.5.4, we can conclude that the null space of \(\matrixOf{T}{B'B}\) is also trivial, in which case \(\matrixOf{T}{B'B}\) is invertible (Statement 9 of Theorem 21.5.5).

Statement 2 implies Statement 3.

This is obvious.

Statement 3 implies Statement 1.

Suppose \(\basisfont{B},\basisfont{B}'\) is a choice of bases for \(V,W\text{,}\) respectively, for which the matrix \(\matrixOf{T}{B'B}\) is square and invertible. Since the dimensions of \(\matrixOf{T}{B'B}\) reflect the dimensions of \(V\) and \(W\text{,}\) these two spaces must have the same dimension. Applying Corollary 44.5.12, it now suffices to verify that \(T\) is injective, which we can do by considering \(\ker T\) (Theorem 44.5.5). But if \(\matrixOf{T}{B'B}\) is invertible, it must have trivial null space (Statement 9 of Theorem 21.5.5), which implies that \(\ker T\) is trivial as well (Statement 1 of Proposition 45.5.4).

We have now completed the cycle of logical dependence to demonstrate that these three statements are equivalent.

Subsection 45.5.3 Matrices of compositions and inverses

As we have thoroughly discussed these relationships in Subsection 45.3.4, we will state them here without proof.

Subsection 45.5.4 The space of linear transformations as a space of matrices

We now have a correspondence that associates a matrix \(\matrixOf{T}{B'B}\) to each linear transformation \(\funcdef{T}{V}{W}\) via a choice of bases \(\basisfont{B},\basisfont{B}'\) of spaces \(V,W\text{,}\) respectively, such that the properties of the transformation are reflected in the properties of the matrix, and vice versa. Essentially, the space of transformations \(L(V,W)\) comes to resemble a space of matrices \(\matrixring_{m \times n}(\R)\) (or \(\matrixring_{m \times n}(\C)\text{,}\) as appropriate).

As usual, we treat only the real case, as the complex case is identical.

Let \(\funcdef{M}{L(V,W)}{\matrixring_{m \times n}(\R)}\) represent the function defined by

\begin{equation*} M(T) = \matrixOf{T}{B'B} \text{.} \end{equation*}

First, we must verify that \(M\) is linear.

Additivity.

Suppose \(T_1,T_2\) are transformations in \(L(V,W)\text{,}\) and let \(\uvec{v}\) be an arbitrary vector in the domain space \(V\text{.}\) Then we have

\begin{align*} (\matrixOf{T_1}{B'B} + \matrixOf{T_2}{B'B}) \matrixOf{\uvec{v}}{B} \amp = \matrixOf{T_1}{B'B} \matrixOf{\uvec{v}}{B} + \matrixOf{T_2}{B'B} \matrixOf{\uvec{v}}{B} \amp \amp\text{(i)}\\ \amp = \matrixOf{T_1(\uvec{v})}{B'} + \matrixOf{T_2(\uvec{v})}{B'} \amp \amp\text{(ii)}\\ \amp = \matrixOf{T_1(\uvec{v}) + T_2(\uvec{v})}{B'} \amp \amp\text{(iii)}\\ \amp = \matrixOf{(T_1 + T_2)(\uvec{v})}{B'} \amp \amp\text{(iv)}\text{,} \end{align*}

with justifications

  1. matrix algebra (Rule 1.d of Proposition 4.5.1);
  2. equality (\(\star\)), applied to both \(T_1\) and \(T_2\text{;}\)
  3. linearity of the coordinate map (Theorem 22.5.1); and
  4. definition of the sum of linear transformations (see Subsection 42.3.6).

But \(\matrixOf{T_1 + T_2}{B'B}\) should be the one unique matrix that satisfies

\begin{equation*} \matrixOf{(T_1 + T_2)(\uvec{v})}{B'} = \matrixOf{T_1 + T_2}{B'B} \matrixOf{\uvec{v}}{B} \end{equation*}

for every vector \(\uvec{v}\) in the domain space \(V\) (Theorem 45.5.1). As the matrix \(\matrixOf{T_1}{B'B} + \matrixOf{T_2}{B'B}\) satisfies the same condition, we must have

\begin{equation*} \matrixOf{T_1 + T_2}{B'B} = \matrixOf{T_1}{B'B} + \matrixOf{T_2}{B'B} \text{.} \end{equation*}

Using the notation of \(\funcdef{M}{L(V,W)}{\matrixring_{m \times n}(\R)}\text{,}\) this says that

\begin{equation*} M(T_1 + T_2) = M(T_1) + M(T_2) \text{,} \end{equation*}

as required.

Homogeneity.

Suppose \(T\) is a transformation in \(L(V,W)\text{,}\) \(k\) is a scalar, and let \(\uvec{v}\) be an arbitrary vector in the domain space \(V\text{.}\) Then we have

\begin{align*} (k \, \matrixOf{T}{B'B}) \matrixOf{\uvec{v}}{B} \amp = k \matrixOf{T}{B'B} \matrixOf{\uvec{v}}{B} \amp \amp\text{(i)}\\ \amp = k \matrixOf{T(\uvec{v})}{B'} \amp \amp\text{(ii)}\\ \amp = \matrixOf{k \, T(\uvec{v})}{B'} \amp \amp\text{(iii)}\\ \amp = \matrixOf{(k T)(\uvec{v})}{B'} \amp \amp\text{(iv)}\text{,} \end{align*}

with justifications

  1. matrix algebra (Rule 2.c of Proposition 4.5.1);
  2. equality (\(\star\));
  3. linearity of the coordinate map (Theorem 22.5.1); and
  4. definition of the scalar multiple of a linear transformation (see Subsection 42.3.6).

But \(\matrixOf{k T}{B'B}\) should be the one unique matrix that satisfies

\begin{equation*} \matrixOf{(k T)(\uvec{v})}{B'} = \matrixOf{k T}{B'B} \matrixOf{\uvec{v}}{B} \end{equation*}

for every vector \(\uvec{v}\) in the domain space \(V\) (Theorem 45.5.1). As the matrix \(k \, \matrixOf{T}{B'B}\) satisfies the same condition, we must have

\begin{equation*} \matrixOf{k T}{B'B} = k \, \matrixOf{T}{B'B} \text{.} \end{equation*}

Using the notation of \(\funcdef{M}{L(V,W)}{\matrixring_{m \times n}(\R)}\text{,}\) this says that

\begin{equation*} M(k T) = k \, M(T) \text{,} \end{equation*}

as required.

Now we check that \(M\) is an isomorphism.

Injectivity.

As we have already shown that \(M\) is linear, we may check that \(M\) is injective by verifying that \(\ker M\) is trivial (Theorem 44.5.5). So suppose that transformation \(T\) in \(L(V,W)\) satisfies \(M(T) = \zerovec_{m \times n}\text{,}\) the \(m \times n\) zero matrix. Then for every \(\uvec{v}\) in \(V\text{,}\) we have

\begin{equation*} \matrixOf{T(\uvec{v})}{B'} = \matrixOf{T}{B'B} \matrixOf{\uvec{v}}{B} = M(T) \matrixOf{\uvec{v}}{B} = \zerovec_{m \times n} \matrixOf{\uvec{v}}{B} = \zerovec_m\text{.} \end{equation*}

But the coordinate map \(\coordmap{B'}\) is an isomorphism, and \(\zerovec_W\) is the one unique vector in \(W\) satisfying

\begin{equation*} \matrixOf{\zerovec_W}{B'} = \zerovec_m \text{,} \end{equation*}

so we must have

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

for each \(\uvec{v}\) in \(V\text{.}\) In particular, this holds for each basis vector in \(\basisfont{B}\text{.}\) However, the zero transformation is the unique transformation \(V \to W\) that sends each vector in the domain space basis \(\basisfont{B}\) to \(\zerovec_W\) (Corollary 42.5.3), and so we can conclude \(T = \zerovec_{V,W}\text{,}\) as desired.

Surjectivity.

We wish to verify that, given matrix \(A\) in \(\matrixring_{m \times n}(\R)\text{,}\) there exists a transformation \(T\) in \(L(V,W)\) with

\begin{equation*} M(T) = A\text{,} \end{equation*}

i.e. with

\begin{equation*} \matrixOf{T}{B'B} = A\text{.} \end{equation*}

As in Corollary 42.5.3, we may attempt to define such a \(T\) by simply specifying the image vectors for each of the domain space basis vectors in \(\basisfont{B}\text{.}\) Writing \(\uvec{a}_1,\dotsc,\uvec{a}_n\) for the columns of \(A\) and \(\uvec{v}_1,\dotsc,\uvec{v}_n\) for the vectors in \(\basisfont{B}\text{,}\) for each index \(j\) set \(T(\uvec{v}_j)\) to be the vector in \(W\) whose coordinate vector in \(\R^m\) is \(\uvec{a}_j\text{.}\) That is, set each

\begin{equation*} T(\uvec{v}_j) = \invcoordmap{B'}(\uvec{a}_j) \text{.} \end{equation*}

Then, using the computation pattern (\(\star\)) developed in Subsection 45.3.2, we see that \(\matrixOf{T}{B'B} = A\) will be true, as desired.

Remark 45.5.8.

Recall that the dual space of a vector space \(V\) is the space \(\vecdual{V} = L(V,\R^1)\) (real case) or \(\vecdual{V} = L(V,\C^1)\) (complex case). Applying the theorem to this situation, we have

\begin{equation*} \vecdual{V} \iso \matrixring_{1 \times n}(\R) \quad \text{or} \quad \vecdual{V} \iso \matrixring_{1 \times n}(\C) \text{,} \end{equation*}

as appropriate, where

\begin{equation*} n = \dim V = \dim \vecdual{V} \text{.} \end{equation*}

This is merely another version of Corollary 44.5.16, where we realize \(\R^n\) and \(\C^n\) as spaces of \(1 \times n\) row vectors instead of \(n \times 1\) column vectors. In other words, in realizing \(V \iso \R^n\) (or \(V \iso \C^n\text{,}\) as appropriate) via coordinate maps, we naturally associate vectors in \(V\) with column vectors, whereas in realizing \(\vecdual{V} \iso \R^n\) (or \(\vecdual{V} \iso \C^n\text{,}\) as appropriate) it is actually more natural to associate vectors in \(\vecdual{V}\) with row vectors.

These associations allow us to think of evaluation of a linear functional \(f\) in \(\vecdual{V}\) on a vector \(\uvec{v}\) in \(V\) as a dot-product-like pairing

\begin{equation*} f(\uvec{v}) = \inprod{\uvec{v}}{f} \text{,} \end{equation*}

so that evaluation \(f(\uvec{v})\) is akin to multiplication of the row vector for \(f\) times the column vector for \(\uvec{v}\text{.}\)