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Section 44.5 Theory

Subsection 44.5.1 Properties of composite transformations

Linearity of \(T\) and \(S\) means that

\begin{align*} T(\uvec{u}_1 + \uvec{u}_2) \amp = T(\uvec{u}_1) + T(\uvec{u}_2) \text{,} \amp S(\uvec{v}_1 + \uvec{v}_2) \amp = S(\uvec{v}_1) + S(\uvec{v}_2) \text{,} \amp\\ T(k \uvec{u}) \amp = k \, T(\uvec{u}) \text{,} \amp S(k \uvec{v}) \amp = k \, S(k \uvec{v}) \amp \end{align*}

for all vectors \(\uvec{u}_1,\uvec{u}_2,\uvec{u}\) in \(U\text{,}\) all vectors \(\uvec{v}_1,\uvec{v}_2,\uvec{v}\) in \(V\text{,}\) and all scalars \(k\text{.}\)

To verify that the composition \(ST\) is linear, the above linearity properties for \(T\) and \(S\) can be used to verify the linearity properties for \(ST\text{:}\)

\begin{align*} ST(\uvec{u}_1 + \uvec{u}_2) \amp = ST(\uvec{u}_1) + ST(\uvec{u}_2) \text{,} \amp ST(k \uvec{u}) \amp = k \, ST(\uvec{u}) \end{align*}

for all vectors \(\uvec{u}_1,\uvec{u}_2,\uvec{u}\) in \(U\text{,}\) and all scalars \(k\text{.}\)

We leave the details to you, the reader.

The linearity of compositions can be extended to the composition of any number of linear transformations, provided domains and codomains match up as necessary to actually form the composition. And compositions of three or more transformations satisfy the associative property.

This is a basic property of compositions of functions, and so holds for linear transformations between vector spaces.

Here are some facts about composition with two special transformations. Verification of these statements is straightforward, and we leave that task to you, the reader.

Finally, we record our observation about the standard matrix of a composition of matrices, without proof.

Subsection 44.5.2 One-to-one and invertible transformations

First, we prove our characterization of injective transformations via the kernel.

Assume \(T\) is injective.

In this case, we would like to verify that \(\ker T = \{\zerovec_V\}\text{.}\) But linear \(T\) must satisfy \(T(\zerovec_V) = \zerovec_W\) (Statement 1 of Proposition 42.5.1). Since \(T\) is injective, there can be no other vectors in the domain space \(V\) with image \(\zerovec_W\text{,}\) and so \(\ker T = \{\zerovec_V\}\text{.}\)

Assume \(T\) has trivial kernel.

That is, we assume \(\ker T = \{\zerovec_V\}\text{,}\) and would like to verify that \(T\) is injective. So suppose \(\uvec{v}_1,\uvec{v}_2\) are vectors in the domain space \(V\) with the same image vector under \(T\text{.}\) Then,

\begin{gather*} \phantom{\implies} T(\uvec{v}_1) = T(\uvec{v}_2) \\ \implies T(\uvec{v}_1) - T(\uvec{v}_2) = \zerovec_W \\ \implies T(\uvec{v}_1 - \uvec{v}_2) = \zerovec_W \text{,} \end{gather*}

where in the last step we have used the linearity of \(T\text{.}\) This final equality says that the difference \(\uvec{v}_1 - \uvec{v}_2\) must be in \(\ker T\text{.}\) But as we have assumed that \(\zerovec_V\) is the only vector in \(\ker T\text{,}\) we have

\begin{gather*} \phantom{\implies} \uvec{v}_1 - \uvec{v}_2 = \zerovec_V \\ \implies \uvec{v}_1 = \uvec{v}_2 \text{.} \end{gather*}

This verifies that different domain space vectors cannot produce the same image vector in the codomain, hence \(T\) is one-to-one.

Through the Dimension Theorem, the above theorem puts a restriction on the dimension of the codomain for a one-to-one transformation.

The Dimension Theorem says that

\begin{equation*} \dim(\ker T) + \dim(\im T) = \dim V \text{.} \end{equation*}

If

\begin{equation*} \dim W \lt \dim V \text{,} \end{equation*}

then also

\begin{equation*} \dim(\im T) \lt \dim V \text{,} \end{equation*}

since \(\im T\) is a subspace of \(W\) (Statement 2 of Proposition 20.5.8). But then

\begin{equation*} \dim(\ker T) = \dim V - \dim(\im T) \gt 0 \text{,} \end{equation*}

so that \(\ker T\) cannot be trivial. The result now follows from Theorem 44.5.5.

As well, if the kernel is trivial, then the transformation must preserve independence. This fact will justify Procedure 44.3.3.

Statement 1 implies Statement 2.

Assume \(S\) is linearly independent in \(V\text{,}\) and suppose \(\uvec{w}_1, \dotsc, \uvec{w}_m\) is a collection of distinct vectors from \(T(S)\text{.}\) Since these are image vectors, there are corresponding vectors \(\uvec{v}_1, \dotsc, \uvec{v}_m\) in \(S\) so that \(T(\uvec{v}_j) = \uvec{w}_j\) for each index \(j\) And since \(T\) is assumed injective, there can be no repeats amongst the \(\uvec{v}_j\text{.}\)

To test the \(\uvec{w}_j\) for linear independence, we follow the Test for Linear Dependence/Independence and form the homogeneous vector equation

\begin{equation*} k_1 \uvec{w}_1 + \dotsb + k_n \uvec{w}_n = \zerovec_W \text{.} \end{equation*}

Making substitutions \(\uvec{w}_j = T(\uvec{v}_j)\) and using the linearity of \(T\text{,}\) we have

\begin{gather*} \phantom{\implies} k_1 T(\uvec{v}_1) + \dotsb + k_n T(\uvec{v}_n) = \zerovec_W \\ \implies T(k_1 \uvec{v}_1 + \dotsb + k_n \uvec{v}_n) = \zerovec_W \text{,} \end{gather*}

which says that

\begin{equation*} k_1 \uvec{v}_1 + \dotsb + k_n \uvec{v}_n \end{equation*}

is in \(\ker T\text{.}\) But we have assumed that \(T\) is injective, so its kernel is trivial (Theorem 44.5.5). Therefore,

\begin{equation*} k_1 \uvec{v}_1 + \dotsb + k_n \uvec{v}_n = \zerovec_V \text{.} \end{equation*}

This is now the beginning of the Test for Linear Dependence/Independence applied to a set of vectors from \(S\text{,}\) which we know is independent. Hence we must have each \(k_j = 0\text{,}\) as desired.

Statement 2 implies Statement 3.

This is obvious from the fact that a basis is linearly independent.

Statement 3 implies Statement 1.

First, we assume that for at least one basis of the domain space \(V\text{,}\) \(T\) sends those basis vectors to a linearly independent set in the codomain space \(W\text{.}\) So let \(\basisfont{B}\) represent just such a basis for \(V\text{.}\)

To verify that \(T\) must be an injective, we can instead verify that \(\ker T\) is trivial (Theorem 44.5.5). So suppose \(\uvec{v}\) is in \(\ker T\text{.}\) We can express \(\uvec{v}\) as a linear combination of \(\basisfont{B}\)-vectors, say

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

From our assumption that \(\uvec{v}\) is in \(\ker T\text{,}\) and using the linearity of \(T\text{,}\) we have

\begin{gather*} \phantom{\implies} T(\uvec{v}) = \zerovec_W \\ \implies T(a_1 \uvec{v}_1 + \dotsb + a_n \uvec{v}_n) = \zerovec_W \\ \implies a_1 T(\uvec{v}_1) + \dotsb + a_n T(\uvec{v}_n) = \zerovec_W \text{.} \end{gather*}

Our assumption about the special basis \(\basisfont{B}\) is that the image vectors \(T(\uvec{v}_j)\) form a linearly independent set in \(W\text{.}\) Hence the only way the last equality above can be true is if each of the \(a_j = 0\text{,}\) which means that

\begin{align*} \uvec{v} \amp = a_1 \uvec{v}_1 + \dotsb + a_n \uvec{v}_n \\ \amp = 0 \uvec{v}_1 + \dotsb + 0 \uvec{v}_n \\ \amp = \zerovec_V \text{.} \end{align*}

We started with the assumption that \(\uvec{v}\) was in \(\ker T\text{,}\) and were able to show that in fact \(\uvec{v} = \zerovec_V\text{.}\) This implies that the only vector in \(\ker T\) is \(\zerovec_V\text{,}\) and so \(\ker T\) is trivial, as required.

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

Having trivial kernel also means that constructing a basis for the image of an injective transformation becomes simpler.

We already know how to construct a basis for \(\im T\) using Theorem 43.5.3. Taking a basis \(\basisfont{K}\) for \(\ker T\) and additional linearly independent vectors \(\basisfont{K}'\) in \(V\) so that together the vectors from \(\basisfont{K}\) and \(\basisfont{K}'\) form a basis for \(V\text{,}\) then the image vectors \(T(\basisfont{K}')\) will form a basis for \(\im T\text{.}\)

However, if \(T\) is injective, then \(\ker T\) is trivial, and so \(\basisfont{K}\) can only be the empty set (see Subsection 20.3.5). Therefore, we can take \(\basisfont{K}'\) to be all of \(\basisfont{B}\text{,}\) and then \(T(\basisfont{K}') = T(\basisfont{B})\) is a basis for \(\im T\text{,}\) as desired.

Now we verify that inverse transformations are, in fact, linear.

Additivity.

Suppose \(\uvec{w}_1,\uvec{w}_2\) are vectors in \(\im T\text{,}\) and set

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

Since \(T\) is injective, these are the unique domain space vectors satisfying

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

But then, 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) = \uvec{w}_1 + \uvec{w}_2 \text{.} \end{equation*}

So \(\uvec{v}_1 + \uvec{v}_2\) must be the unique domain space vector that produces image vector \(\uvec{w}_1 + \uvec{w}_2\text{.}\) In other words,

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

as required.

Homogeneity.

Suppose \(\uvec{w}\) is a vector in \(\im T\text{,}\) and set

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

Since \(T\) is injective, this is the unique domain space vectors satisfying

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

But then, using the Homogeneity of \(T\text{,}\) for each scalar \(k\) we have

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

So \(k \uvec{v}\) must be the unique domain space vector that produces image vector \(k \uvec{w}\text{.}\) In other words,

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

as required.

The purpose of an inverse function is to reverse the original function, and that remains the case for inverses of linear transformations.

  1. For \(\uvec{v}\) in \(V\text{,}\) write \(\uvec{w} = T(\uvec{v})\text{.}\) By definition, \(\inv{T}(\uvec{w})\) is the unique vector in the domain space \(V\) whose image vector under \(T\) is \(\uvec{w}\text{.}\) Clearly, that domain space vector is \(\uvec{v}\text{.}\)
  2. For \(\uvec{w}\) in \(\im T\text{,}\) write \(\uvec{v} = \inv{T}(\uvec{w})\text{.}\) By definition, \(\uvec{v}\) is the unique vector in the domain space \(V\) satisfying \(T(\uvec{v}) = \uvec{w}\text{.}\) But then this says \((T \inv{T}) (\uvec{w}) = \uvec{w}\text{,}\) as desired.

Subsection 44.5.3 Isomorphisms

An isomorphism is a linear transformation that is both injective and surjective. We have already characterized injective transformations above in terms of the kernel (Theorem 44.5.5), and the concept of surjective is defined in terms of the image. But here we will characterize surjectivity in terms of the dimensions of those spaces.

Recall that, by definition, \(T\) is surjective if \(\im T = W\text{.}\) Since \(\im T\) is a subspace of \(W\text{,}\) then the first statement follows from Statement 3 of Proposition 20.5.8. And the second statement follows from the first via the Dimension Theorem.

Similarly to how dimension lets us reduce the task of verifying a basis to one of linear independence and spanning (Corollary 20.5.6), dimension also lets us reduce checking isomorphism to one of injective and surjective.

Assume \(T\) is injective.

Then \(\nullity T = 0\text{,}\) hence \(\rank T = \dim V\) by the Dimension Theorem. But we have assumed \(\dim V = \dim W\text{,}\) so we have \(\rank T = \dim W\text{,}\) hence \(T\) is surjective by Statement 1 of Proposition 44.5.11.

Assume \(T\) is surjective.

Hence \(\rank T = \dim W\) (Statement 1 of Proposition 44.5.11). But we have assumed \(\dim W = \dim V\text{,}\) so we have \(\rank T = \dim V\) as well. By the Dimension Theorem, we must have \(\nullity T = 0\text{,}\) so that \(\ker T\) is trivial. Therefore, \(T\) is injective, as desired (Theorem 44.5.5),

Before we characterize isomorphic spaces in terms of dimension, we will first characterize isomorphism in terms of basis. (Which makes sense, since dimension is defined in terms of basis.) This characterization will also justify Procedure 44.3.4.

Statement 1 implies Statement 2.

This is essentially just Corollary 44.5.8, using the fact that for an isomorphism \(\funcdef{T}{V}{W}\) we have \(\im T = W\text{.}\)

Statement 2 implies Statement 3.

This is obvious.

Statement 3 implies Statement 1.

First, we assume that for at least one basis of the domain space \(V\text{,}\) \(T\) sends those basis vectors to a basis for the codomain space \(W\text{.}\) So let \(\basisfont{B}\) represent just such a basis for \(V\text{.}\) Then \(T(\basisfont{B})\) cannot contain any repeat image vectors, since it is a linearly independent set. Hence \(\basisfont{B}\) and \(T(\basisfont{B})\) must have the same number of vectors. In other words, \(\dim V = \dim W\text{.}\)

With this established, we can use Corollary 44.5.12 to verify that \(T\) must be an isomorphism by checking that \(T\) is injective. But injectivity is immediate using the equivalence of Statement 1 and Statement 3 of Corollary 44.5.7, given that every basis is linearly independent.

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

Remark 44.5.14.

It may seem that including Statement 3 of the theorem is redundant, given Statement 2. However, the direction of logic is important when applying this theorem. If we already know (or assume) that a transformation \(T\) is an isomorphism, then we would like as strong a statement as possible about what \(T\) does to bases of the domain space \(V\text{,}\) hence the “every” in Statement 2. On the other hand, if we are trying to verify that a transformation \(T\) is an isomorphism, then we would like as weak a condition as possible to reduce the amount of effort necessary, hence the “at least one” in Statement 3.

The theorem says that all we need to do to create an isomorphism is create a correspondence between bases. The only potential barrier to carrying this out is if the number of vectors in bases for domain and codomain spaces do not match.

Assume \(V,W\) are isomorphic.

By definition, this means that there exists some isomorphism \(\funcdef{T}{V}{W}\text{.}\) By the equivalence of the statements of Theorem 44.5.13, \(T\) sends a basis for \(V\) to a basis for \(W\text{.}\) But then these two bases have the same number of vectors, and so the dimensions of the two spaces are equal.

Assume \(V,W\) have the same dimension.

Choose bases of \(V\) and \(W\text{:}\)

\begin{align*} \basisfont{B}_V \amp = \{\uvec{v}_1,\dotsc,\uvec{v}_n\} \text{,} \\ \basisfont{B}_W \amp = \{\uvec{w}_1,\dotsc,\uvec{w}_n\} \text{.} \end{align*}

Let \(\funcdef{T}{V}{W}\) be the unique linear transformation satisfying \(T(\uvec{v}_j) = \uvec{w}_j\) for each index \(j\) (Corollary 42.5.3). But then the equivalence of Statement 1 and Statement 3 of Theorem 44.5.13 tells us that \(T\) is an ismorphism, which verifies that \(V,W\) are isomorphic.

In each of the real and complex cases we have a favourite \(n\)-dimensional space to which others must be isomorphic.

We have already seen how simply specifying a basis for an \(n\)-dimensional vector space creates a very specific isomorphism to \(\R^n\) or \(\C^n\) (real or complex case).

In Subsection 44.3.9, we justified the claim in the statement of this corollary by appealing to kernel and image, but Theorem 44.5.13 gives one simple criterion to verify. The collection \(\coordmap{B}(\basisfont{B})\) of image vectors for the specific basis \(\basisfont{B}\) of \(V\) used to create the coordinate map \(\coordmap{B}\) is particularly simple to compute. If we write

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

then from

\begin{equation*} \uvec{v}_j = 0 \uvec{v}_1 + \dotsb + 0 \uvec{v}_{j-1} + 1 \uvec{v}_j + 0 \uvec{v}_{j+1} + \dotsb + 0 \uvec{v}_n \end{equation*}

we have

\begin{equation*} \coordmap{B}(\uvec{v}_j) = (0,\dotsc,0,1,0,\dotsc,0) \end{equation*}

in \(\R^n\) or \(\C^n\text{,}\) as appropriate. That is, the collection \(\coordmap{B}(\basisfont{B})\) of image vectors is actually the collection of standard basis vectors in \(\R^n\) or \(\C^n\text{,}\) as appropriate, so Statement 3 of Theorem 44.5.13 is satisfied by \(\coordmap{B}\text{.}\)

Remark 44.5.18.

In fact, if you look back at the proof of Corollary 44.5.15, the isomorphism constructed in the second case by sending a basis of \(V\) to a basis of \(W\) is precisely the coordinate map relative to the basis for \(V\) in the case that \(W\) is \(\R^n\) (real case) or \(\C^n\) (complex case), and the basis for \(W\) is chosen to be the standard basis.

Combining Corollary 44.5.15 and Corollary 42.5.8 leads to the following.

Remark 44.5.20.

Via Theorem 44.5.13, Theorem 42.5.7 provides a means to realizing an isomorphism \(V \to \vecdual{V}\) — choose a basis for \(V\) and send each of those basis vectors to the corresponding vector in the dual basis provided by that theorem.

By the same reasoning, every finite-dimensional vector space is also isomorphic to its double dual space, since all three spaces \(V, \vecdual{V}, \vecddual{V}\) always have the same dimension as each other. However, creating an isomorphism between a space \(V\) and its dual space \(\vecdual{V}\) always requires first choosing a basis for \(V\text{.}\) The isomorphic relationship between \(V\) and its double dual space \(\vecddual{V}\) is special in that a coordinate-free isomorphism can be created, as in the following theorem.

  1. It must be checked that each \(F_{\uvec{v}}\) satisfies the linearity properties: additivity and homogeneity. We leave this task to you, the reader.

  2. From Corollary 42.5.8, we may conclude that \(\dim \vecddual{V} = \dim V\text{.}\) So by Corollary 44.5.12, it suffices to check that \(E\) is injective. And for this task it suffices to check that \(\ker E\) is trivial, by Theorem 44.5.5.

    So suppose that vector \(\uvec{v}\) in \(V\) is in \(\ker E\text{,}\) so that

    \begin{equation*} E(\uvec{v}) = \zerovec_{\vecddual{V}} \text{,} \end{equation*}

    the zero functional on \(\vecdual{V}\text{.}\) Then for each element \(f\) in \(\vecdual{V}\) (i.e. for each linear functional \(f\) on \(V\)), we have

    \begin{equation*} E(\uvec{v})(f) = \zerovec_{\vecddual{V}}(f) = 0 \text{.} \end{equation*}

    But by our definitions, we have

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

    so we may conclude that

    \begin{equation*} f(\uvec{v}) = 0 \end{equation*}

    for each linear functional \(f\) on \(V\text{.}\) If \(\uvec{v}\) is not the zero vector of \(V\text{,}\) then the linearly independent set \(\{\uvec{v}\}\) can be enlarged to a basis for \(V\) (Proposition 20.5.4), from which we can obtain a dual basis for \(\vecdual{V}\) (Theorem 42.5.7). But then the first vector in this dual basis will be \(\vecdual{\uvec{v}}\text{,}\) for which

    \begin{equation*} \vecdual{\uvec{v}}(\uvec{v}) = 1 \text{,} \end{equation*}

    which contradicts the conclusion that every linear functional on \(V\) evaluates to zero on \(\uvec{v}\text{.}\) So the only possibility for a vector \(\uvec{v}\) to be in \(\ker E\) is \(\uvec{v} = \zerovec_V\text{.}\) That is, \(\ker E\) is trivial, as desired.

Theorem 44.5.13 also has immediate consequences for the inverse of an isomorphism and the composition of isomorphisms.

We will make repeated use of the equivalence of the statements of Theorem 44.5.13, first applied to \(T\) and then to \(\inv{T}\text{.}\)

Suppose \(\basisfont{B}\) is a basis for the domain space \(V\text{.}\) Because \(T\) is assumed to be an isomorphism, the collection \(T(\basisfont{B})\) of image vectors for the vectors in \(\basisfont{B}\) is a basis for \(W\text{.}\) But clearly \(\inv{T}\) sends the vectors of \(T(\basisfont{B})\) right back to their counterparts in \(\basisfont{B}\text{.}\) That is, \(\inv{T}\) sends at least one basis for \(W\) to a basis for \(V\text{,}\) and hence must be an isomorphism.

Now, \(T\) sends each vector in \(\basisfont{B}\) to its image in \(T(\basisfont{B})\text{,}\) and \(\inv{T}\) sends each vector in \(T(\basisfont{B})\) right back. So \(\inv{(\inv{T})}\) must send the vectors in \(\basisfont{B}\) forward again to their corresponding \(T\)-images. But this says that \(T\) and \(\inv{(\inv{T})}\) act the same on basis \(\basisfont{B}\text{,}\) which implies that they must be the same transformation (Theorem 42.5.2).

Again, we will make repeated use of the equivalence of the statements of Theorem 44.5.13, first applied to \(T\) and \(S\text{,}\) and then to the composition \(ST\text{.}\)

Suppose \(\basisfont{B}_U\) is a basis for the domain space \(U\) of \(T\text{.}\) Because \(T\) is assumed to be an isomorphism, the collection \(\basisfont{B}_V = T(\basisfont{B}_U)\) of image vectors for the vectors in \(\basisfont{B}_U\) is a basis for \(V\text{.}\) And then also the collection \(\basisfont{B}_W = S(\basisfont{B}_V)\) of image vectors for the vectors in \(\basisfont{B}_V\) is a basis for \(W\text{,}\) because \(S\) is also assumed to be an isomorphism. But then

\begin{equation*} \basisfont{B}_W = S\bigl(T(\basisfont{B}_U)\bigr) = ST(\basisfont{B}_U) \text{,} \end{equation*}

so that the composition \(ST\) takes basis \(\basisfont{B}_U\) for its domain to a basis \(\basisfont{B}_W\) for its codomain. Applying the theorem, we may conclude that \(ST\) is an isomorphism.

Now, \(T\) sends each vector in \(\basisfont{B}_U\) to its image in \(\basisfont{B}_V\text{,}\) and \(\inv{T}\) sends each vector in \(\basisfont{B}_V\) right back. Similarly, \(S\) sends each vector in \(\basisfont{B}_V\) to its image in \(\basisfont{B}_W\text{,}\) and \(\inv{S}\) sends each vector in \(\basisfont{B}_W\) right back. Chaining this forward, we can say that \(ST\) sends each vector in \(\basisfont{B}_U\) to the corresponding vector in \(\basisfont{B}_W\text{.}\) But we can also trace this backwards to see that both \(\inv{(S T)}\) and \(\inv{T} \inv{S}\) send each vector in \(\basisfont{B}_W\) to its counterpart in \(\basisfont{B}_U\text{.}\) This says that \(\inv{(S T)}\) and \(\inv{T} \inv{S}\) act the same on basis \(\basisfont{B}_W\text{,}\) from which we may conclude that they are the same transformation (Theorem 42.5.2).

Finally, we record our observation about invertibility of standard matrices of transformations \(\R^n \to \R^n\text{.}\)

We will prove only the real case; the complex case is identical.

Assume \(T\) is an isomorphism.

In this case, \(T\) sends each basis for the domain space \(\R^n\) to a basis for the codomain space \(\R^m\) (Theorem 44.5.13). Therefore, these two spaces must have the same dimension, i.e. \(m = n\text{.}\) And, in particular, \(T\) must send the standard basis of \(\R^n\) to another basis for \(\R^n\text{.}\) But

\begin{equation*} T(\uvec{e}_j) = \stdmatrixOf{T} \uvec{e}_j \end{equation*}

results in the \(\nth[j]\) column of \(\stdmatrixOf{T}\text{,}\) so the columns of \(\stdmatrixOf{T}\) must form a basis of \(\R^n\text{.}\) From this, we can conclude that \(\stdmatrixOf{T}\) must be invertible using Statement 11 of Theorem 21.5.5.

Assume \(m = n\) and that \(\stdmatrixOf{T}\) is an invertible matrix.

Since we assume that domain and codomain spaces have the same dimension, we can use Corollary 44.5.12 to verify that \(T\) must be an isomorphism by checking that \(T\) is injective. And to do that, we can check that \(\ker T\) is trivial (Theorem 44.5.5). But for matrix transformation \(T\text{,}\) \(\ker T\) is the same as the null space of \(\stdmatrixOf{T}\text{.}\) So using Statement 9 of Theorem 21.5.5, we can conclude that \(\ker T\) is trivial from our assumption that \(\stdmatrixOf{T}\) is invertible.

It remains to verify that \(\stdmatrixOf{\inv{T}} = \inv{\stdmatrixOf{T}} \text{.}\) However, using Proposition 44.5.4 and Proposition 44.5.10, we know that

\begin{equation*} \stdmatrixOf{T} \stdmatrixOf{\inv{T}} = \stdmatrixOf{T \inv{T}} = \stdmatrixOf{I_{\R^n}} = I\text{,} \end{equation*}

the \(n \times n\) identity matrix. We can thus conclude that \(\stdmatrixOf{\inv{T}} \) and \(\stdmatrixOf{T}\) are inverses of each other, using Proposition 6.5.4.