##### DefinitionIDLTIdentity Linear Transformation

The *identity linear transformation* on the vector space \(W\) is defined as
\begin{equation*}
\ltdefn{I_W}{W}{W},\quad\quad \lteval{I_W}{\vect{w}}=\vect{w}\text{.}
\end{equation*}

In this section we will conclude our introduction to linear transformations by bringing together the twin properties of injectivity and surjectivity and consider linear transformations with both of these properties.

One preliminary definition, and then we will have our main definition for this section.

The *identity linear transformation* on the vector space \(W\) is defined as
\begin{equation*}
\ltdefn{I_W}{W}{W},\quad\quad \lteval{I_W}{\vect{w}}=\vect{w}\text{.}
\end{equation*}

Informally, \(I_W\) is the “do-nothing” function. You should check that \(I_W\) is really a linear transformation, as claimed, and then compute its kernel and range to see that it is both injective and surjective. All of these facts should be straightforward to verify (Exercise IVLT.T05). With this in hand we can make our main definition.

Suppose that \(\ltdefn{T}{U}{V}\) is a linear transformation. If there is a function \(\ltdefn{S}{V}{U}\) such that
\begin{align*}
\compose{S}{T}&=I_U & \compose{T}{S}&=I_V
\end{align*}
then \(T\) is *invertible*. In this case, we call \(S\) the *inverse* of \(T\) and write \(S=\ltinverse{T}\text{.}\)

Informally, a linear transformation \(T\) is invertible if there is a companion linear transformation, \(S\text{,}\) which “undoes” the action of \(T\text{.}\) When the two linear transformations are applied consecutively (composition), in either order, the result is to have no real effect. It is entirely analogous to squaring a positive number and then taking its (positive) square root.

Here is an example of a linear transformation that is invertible. As usual at the beginning of a section, do not be concerned with where \(S\) came from, just understand how it illustrates Definition IVLT.

It can be as instructive to study a linear transformation that is not invertible.

In Example ANILT you may have noticed that \(T\) is not surjective, since the matrix \(A\) was not in the range of \(T\text{.}\) And \(T\) is not injective since there are two different input column vectors that \(T\) sends to the matrix \(B\text{.}\) Linear transformations \(T\) that are not surjective lead to putative inverse functions \(S\) that are undefined on inputs outside of the range of \(T\text{.}\) Linear transformations \(T\) that are not injective lead to putative inverse functions \(S\) that are multiply-defined on each of their inputs. We will formalize these ideas in Theorem ILTIS.

But first notice in Definition IVLT that we only require the inverse (when it exists) to be a function. When it does exist, it too is a linear transformation.

Suppose that \(\ltdefn{T}{U}{V}\) is an invertible linear transformation. Then the function \(\ltdefn{\ltinverse{T}}{V}{U}\) is a linear transformation.

So when \(T\) has an inverse, \(\ltinverse{T}\) is also a linear transformation. Furthermore, \(\ltinverse{T}\) is an invertible linear transformation and *its* inverse is what you might expect.

Suppose that \(\ltdefn{T}{U}{V}\) is an invertible linear transformation. Then \(\ltinverse{T}\) is an invertible linear transformation and \(\ltinverse{\left(\ltinverse{T}\right)}=T\text{.}\)

We now know what an inverse linear transformation is, but just which linear transformations have inverses? Here is a theorem we have been preparing for all chapter long.

Suppose \(\ltdefn{T}{U}{V}\) is a linear transformation. Then \(T\) is invertible if and only if \(T\) is injective and surjective.

When a linear transformation is both injective and surjective, the pre-image of any element of the codomain is a set of size one (a “singleton”). This fact allowed us to *construct* the inverse linear transformation in one half of the proof of Theorem ILTIS (see Proof Technique C) and is illustrated in the following cartoon. This should remind you of the very general Figure 7.48 which was used to illustrate Theorem KPI about pre-images, only now we have an invertible linear transformation which is therefore surjective and injective (Theorem ILTIS). As a surjective linear transformation, there are no vectors depicted in the codomain, \(V\text{,}\) that have empty pre-images. More importantly, as an injective linear transformation, the kernel is trivial (Theorem KILT), so each pre-image is a single vector. This makes it possible to “turn around” all the arrows to create the inverse linear transformation \(\ltinverse{T}\text{.}\)

Many will call an injective and surjective function a *bijective* function or just a *bijection*. Theorem ILTIS tells us that this is just a synonym for the term invertible (which we will use exclusively).

We can follow the constructive approach of the proof of Theorem ILTIS to construct the inverse of a specific linear transformation, as the next example shows.

We will make frequent use of the characterization of invertible linear transformations provided by Theorem ILTIS. The next theorem is a good example of this, and we will use it often, too.

Suppose that \(\ltdefn{T}{U}{V}\) and \(\ltdefn{S}{V}{W}\) are invertible linear transformations. Then the composition, \(\ltdefn{\left(\compose{S}{T}\right)}{U}{W}\) is an invertible linear transformation.

When a composition is invertible, the inverse is easy to construct.

Suppose that \(\ltdefn{T}{U}{V}\) and \(\ltdefn{S}{V}{W}\) are invertible linear transformations. Then \(\compose{S}{T}\) is invertible and \(\ltinverse{\left(\compose{S}{T}\right)}=\compose{\ltinverse{T}}{\ltinverse{S}}\text{.}\)

Notice that this theorem not only establishes *what* the inverse of \(\compose{S}{T}\) *is*, it also duplicates the conclusion of Theorem CIVLT and also establishes the invertibility of \(\compose{S}{T}\text{.}\) But somehow, the proof of Theorem CIVLT is a nicer way to get this property.

Does Theorem ICLT remind you of the flavor of any theorem we have seen about matrices? (Hint: Think about getting dressed.) Hmmmm.

A vector space is defined (Definition VS) as a set of objects (“vectors”) endowed with a definition of vector addition (\(+\)) and a definition of scalar multiplication (written with juxtaposition). Many of our definitions about vector spaces involve linear combinations (Definition LC), such as the span of a set (Definition SS) and linear independence (Definition LI). Other definitions are built up from these ideas, such as bases (Definition B) and dimension (Definition D). The defining properties of a linear transformation require that a function “respect” the operations of the two vector spaces that are the domain and the codomain (Definition LT). Finally, an invertible linear transformation is one that can be “undone” — it has a companion that reverses its effect. In this subsection we are going to begin to roll all these ideas into one.

A vector space has “structure” derived from definitions of the two operations and the requirement that these operations interact in ways that satisfy the ten properties of Definition VS. When two different vector spaces have an invertible linear transformation defined between them, then we can translate questions about linear combinations (spans, linear independence, bases, dimension) from the first vector space to the second. The answers obtained in the second vector space can then be translated back, via the inverse linear transformation, and interpreted in the setting of the first vector space. We say that these invertible linear transformations “preserve structure.” And we say that the two vector spaces are “structurally the same.” The precise term is “isomorphic,” from Greek meaning “of the same form.” Let us begin to try to understand this important concept.

Two vector spaces \(U\) and \(V\) are *isomorphic* if there exists an invertible linear transformation \(T\) with domain \(U\) and codomain \(V\text{,}\) \(\ltdefn{T}{U}{V}\text{.}\) In this case, we write \(U\isomorphic V\text{,}\) and the linear transformation \(T\) is known as an *isomorphism* between \(U\) and \(V\text{.}\)

A few comments on this definition. First, be careful with your language (Proof Technique L). Two vector spaces are isomorphic, or not. It is a yes/no situation and the term only applies to a pair of vector spaces. Any invertible linear transformation can be called an isomorphism, it is a term that applies to functions. Second, given a pair of vector spaces there might be several different isomorphisms between the two vector spaces. But it only takes the existence of one to call the pair isomorphic. Third, \(U\) isomorphic to \(V\text{,}\) or \(V\) isomorphic to \(U\text{?}\) It does not matter, since the inverse linear transformation will provide the needed isomorphism in the “opposite” direction. Being “isomorphic to” is an equivalence relation on the set of all vector spaces (see Theorem SER for a reminder about equivalence relations).

In Example IVSAV we avoided a computation in \(P_3\) by a conversion of the computation to a new vector space, \(M_{22}\text{,}\) via an invertible linear transformation (also known as an isomorphism). Here is a diagram meant to illustrate the more general situation of two vector spaces, \(U\) and \(V\text{,}\) and an invertible linear transformation, \(T\text{.}\) The diagram is simply about a sum of two vectors from \(U\text{,}\) rather than a more involved linear combination. It should remind you of Figure 7.2.

To understand this diagram, begin in the upper-left corner, and by going straight down we can compute the sum of the two vectors using the addition for the vector space \(U\text{.}\) The more circuitous alternative, in the spirit of Example IVSAV, is to begin in the upper-left corner and then proceed clockwise around the other three sides of the rectangle. Notice that the vector addition is accomplished using the addition in the vector space \(V\text{.}\) Then, because \(T\) is a linear transformation, we can say that the result of \(\lteval{T}{\vect{u}_1}+\lteval{T}{\vect{u}_2}\) is equal to \(\lteval{T}{\vect{u}_1+\vect{u}_2}\text{.}\) Then the key feature is to recognize that applying \(\ltinverse{T}\) obviously converts the second version of this result into the sum in the lower-left corner. So there are two routes to the sum \(\vect{u}_1+\vect{u}_2\text{,}\) each employing an addition from a different vector space, but one is “direct” and the other is “roundabout”. You might try designing a similar diagram for the case of scalar multiplication (see Figure 7.3) or for a full linear combination.

Checking the dimensions of two vector spaces can be a quick way to establish that they are not isomorphic. Here is the theorem.

Suppose \(U\) and \(V\) are isomorphic vector spaces. Then \(\dimension{U}=\dimension{V}\text{.}\)

The contrapositive of Theorem IVSED says that if \(U\) and \(V\) have different dimensions, then they are not isomorphic. Dimension is the simplest “structural” characteristic that will allow you to distinguish non-isomorphic vector spaces. For example \(P_6\) is not isomorphic to \(M_{34}\) since their dimensions (7 and 12, respectively) are not equal. With tools developed in Section VR we will be able to establish that the converse of Theorem IVSED is true. Think about that one for a moment.

Just as a matrix has a rank and a nullity, so too do linear transformations. And just like the rank and nullity of a matrix are related (they sum to the number of columns, Theorem RPNC) the rank and nullity of a linear transformation are related. Here are the definitions and theorems, see the Archetypes (Appendix A) for loads of examples.

Suppose that \(\ltdefn{T}{U}{V}\) is a linear transformation. Then the *rank* of \(T\text{,}\) \(\rank{T}\text{,}\) is the dimension of the range of \(T\text{,}\)
\begin{equation*}
\rank{T}=\dimension{\rng{T}}\text{.}
\end{equation*}

Suppose that \(\ltdefn{T}{U}{V}\) is a linear transformation. Then the *nullity* of \(T\text{,}\) \(\nullity{T}\text{,}\) is the dimension of the kernel of \(T\text{,}\)
\begin{equation*}
\nullity{T}=\dimension{\krn{T}}\text{.}
\end{equation*}

Here are two quick theorems.

Suppose that \(\ltdefn{T}{U}{V}\) is a linear transformation. Then the rank of \(T\) is the dimension of \(V\text{,}\) \(\rank{T}=\dimension{V}\text{,}\) if and only if \(T\) is surjective.

Suppose that \(\ltdefn{T}{U}{V}\) is a linear transformation. Then the nullity of \(T\) is zero, \(\nullity{T}=0\text{,}\) if and only if \(T\) is injective.

Just as injectivity and surjectivity come together in invertible linear transformations, there is a clear relationship between rank and nullity of a linear transformation. If one is big, the other is small.

Suppose that \(\ltdefn{T}{U}{V}\) is a linear transformation. Then \begin{equation*} \rank{T}+\nullity{T}=\dimension{U}\text{.} \end{equation*}

Theorem RPNC said that the rank and nullity of a matrix sum to the number of columns of the matrix. This result is now an easy consequence of Theorem RPNDD when we consider the linear transformation \(\ltdefn{T}{\complex{n}}{\complex{m}}\) defined with the \(m\times n\) matrix \(A\) by \(\lteval{T}{\vect{x}}=A\vect{x}\text{.}\) The range and kernel of \(T\) are identical to the column space and null space of the matrix \(A\) (Exercise ILT.T20, Exercise SLT.T20), so the rank and nullity of the matrix \(A\) are identical to the rank and nullity of the linear transformation \(T\text{.}\) The dimension of the domain of \(T\) is the dimension of \(\complex{n}\text{,}\) exactly the number of columns for the matrix \(A\text{.}\)

This theorem can be especially useful in determining basic properties of linear transformations. For example, suppose that \(\ltdefn{T}{\complex{6}}{\complex{6}}\) is a linear transformation and you are able to quickly establish that the kernel is trivial. Then \(\nullity{T}=0\text{.}\) First this means that \(T\) is injective by Theorem NOILT. Also, Theorem RPNDD becomes \begin{equation*} 6=\dimension{\complex{6}}=\rank{T}+\nullity{T}=\rank{T}+0=\rank{T}\text{.} \end{equation*} So the rank of \(T\) is equal to the dimension of the codomain, and by Theorem ROSLT we know \(T\) is surjective. Finally, we know \(T\) is invertible by Theorem ILTIS. So from the determination that the kernel is trivial, and consideration of various dimensions, the theorems of this section allow us to conclude the existence of an inverse linear transformation for \(T\text{.}\)

Similarly, Theorem RPNDD can be used to provide alternative proofs for Theorem ILTD, Theorem SLTD and Theorem IVSED. It would be an interesting exercise to construct these proofs.

It would be instructive to study the archetypes that are linear transformations and see how many of their properties can be deduced just from considering only the dimensions of the domain and codomain. Then add in just knowledge of either the nullity or rank, and see how much more you can learn about the linear transformation. The table preceding all of the archetypes (Appendix A) could be a good place to start this analysis.

This subsection does not really belong in this section, or any other section, for that matter. It is just the right time to have a discussion about the connections between the central topic of linear algebra, linear transformations, and our motivating topic from Chapter SLE, systems of linear equations. We will discuss several theorems we have seen already, but we will also make some forward-looking statements that will be justified in Chapter R.

Archetype D and Archetype E are ideal examples to illustrate connections with linear transformations. Both have the same coefficient matrix, \begin{equation*} D=\begin{bmatrix} 2 & 1 & 7 & -7\\ -3 & 4 & -5 & -6\\ 1 & 1 & 4 & -5 \end{bmatrix}\text{.} \end{equation*}

To apply the *theory* of linear transformations to these two archetypes, employ the matrix-vector product (Definition MVP) and define the linear transformation,
\begin{equation*}
\ltdefn{T}{\complex{4}}{\complex{3}},\quad \lteval{T}{\vect{x}}=D\vect{x}
=x_1\colvector{2\\-3\\1}+
x_2\colvector{1\\4\\1}+
x_3\colvector{7\\-5\\4}+
x_4\colvector{-7\\-6\\-5}\text{.}
\end{equation*}

Theorem MBLT tells us that \(T\) is indeed a linear transformation. Archetype D asks for solutions to \(\linearsystem{D}{\vect{b}}\text{,}\) where \(\vect{b}=\colvector{8\\-12\\-4}\text{.}\) In the language of linear transformations this is equivalent to asking for \(\preimage{T}{\vect{b}}\text{.}\) In the language of vectors and matrices it asks for a linear combination of the four columns of \(D\) that will equal \(\vect{b}\text{.}\) One solution listed is \(\vect{w}=\colvector{7\\8\\1\\3}\text{.}\) With a nonempty preimage, Theorem KPI tells us that the complete solution set of the linear system is the preimage of \(\vect{b}\text{,}\) \begin{equation*} \vect{w}+\krn{T}=\setparts{\vect{w}+\vect{z}}{\vect{z}\in\krn{T}}\text{.} \end{equation*}

The kernel of the linear transformation \(T\) is exactly the null space of the matrix \(D\) (see Exercise ILT.T20), so this approach to the solution set should be reminiscent of Theorem PSPHS. The kernel of the linear transformation is the preimage of the zero vector, exactly equal to the solution set of the homogeneous system \(\homosystem{D}\text{.}\) Since \(D\) has a null space of dimension two, every preimage (and in particular the preimage of \(\vect{b}\)) is as “big” as a subspace of dimension two (but is not a subspace).

Archetype E is identical to Archetype D but with a different vector of constants, \(\vect{d}=\colvector{2\\3\\2}\text{.}\) We can use the same linear transformation \(T\) to discuss this system of equations since the coefficient matrix is identical. Now the set of solutions to \(\linearsystem{D}{\vect{d}}\) is the pre-image of \(\vect{d}\text{,}\) \(\preimage{T}{\vect{d}}\text{.}\) However, the vector \(\vect{d}\) is not in the range of the linear transformation (nor is it in the column space of the matrix, since these two sets are equal by Exercise SLT.T20). So the empty pre-image is equivalent to the inconsistency of the linear system.

These two archetypes each have three equations in four variables, so either the resulting linear systems are inconsistent, or they are consistent and application of Theorem CMVEI tells us that the system has infinitely many solutions. Considering these same parameters for the linear transformation, the dimension of the domain, \(\complex{4}\text{,}\) is four, while the codomain, \(\complex{3}\text{,}\) has dimension three. Then \begin{align*} \nullity{T}&=\dimension{\complex{4}}-\rank{T}&& \knowl{./knowl/theorem-RPNDD.html}{\text{Theorem RPNDD}}\\ &=4-\dimension{\rng{T}}&& \knowl{./knowl/definition-ROLT.html}{\text{Definition ROLT}}\\ &\geq 4-3&& \rng{T}\text{ subspace of }\complex{3}\\ &=1\text{.} \end{align*}

So the kernel of \(T\) is nontrivial simply by considering the dimensions of the domain (number of variables) and the codomain (number of equations). Pre-images of elements of the codomain that are not in the range of \(T\) are empty (inconsistent systems). For elements of the codomain that are in the range of \(T\) (consistent systems), Theorem KPI tells us that the pre-images are built from the kernel, and with a nontrivial kernel, these pre-images are infinite (infinitely many solutions).

When do systems of equations have unique solutions? Consider the system of linear equations \(\linearsystem{C}{\vect{f}}\) and the linear transformation \(\lteval{S}{\vect{x}}=C\vect{x}\text{.}\) If \(S\) has a trivial kernel, then pre-images will either be empty or be finite sets with single elements. Correspondingly, the coefficient matrix \(C\) will have a trivial null space and solution sets will either be empty (inconsistent) or contain a single solution (unique solution). Should the matrix be square and have a trivial null space then we recognize the matrix as being nonsingular. A square matrix means that the corresponding linear transformation, \(T\text{,}\) has equal-sized domain and codomain. With a nullity of zero, \(T\) is injective, and also Theorem RPNDD tells us that rank of \(T\) is equal to the dimension of the domain, which in turn is equal to the dimension of the codomain. In other words, \(T\) is surjective. Injective and surjective, and Theorem ILTIS tells us that \(T\) is invertible. Just as we can use the inverse of the coefficient matrix to find the unique solution of any linear system with a nonsingular coefficient matrix (Theorem SNCM), we can use the inverse of the linear transformation to construct the unique element of any pre-image (proof of Theorem ILTIS).

The executive summary of this discussion is that to every coefficient matrix of a system of linear equations we can associate a natural linear transformation. Solution sets for systems with this coefficient matrix are preimages of elements of the codomain of the linear transformation. For every theorem about systems of linear equations there is an analogue about linear transformations. The theory of linear transformations provides all the tools to recreate the theory of solutions to linear systems of equations.

We will continue this adventure in Chapter R.

What conditions allow us to easily determine if a linear transformation is invertible?

What does it mean to say two vector spaces are isomorphic? Both technically, and informally?

How do linear transformations relate to systems of linear equations?

The archetypes below are linear transformations of the form \(\ltdefn{T}{U}{V}\) that are invertible. For each, the inverse linear transformation is given explicitly as part of the archetype's description. Verify for each linear transformation that \begin{align*} \compose{\ltinverse{T}}{T}&=I_U & \compose{T}{\ltinverse{T}}&=I_V \end{align*} Archetype R, Archetype V, Archetype W

Determine if the linear transformation \(\ltdefn{T}{P_2}{M_{22}}\) is (a) injective, (b) surjective, (c) invertible. \begin{equation*} \lteval{T}{a+bx+cx^2}= \begin{bmatrix} a+2b-2c & 2a+2b \\ -a+b-4c & 3a+2b+2c \end{bmatrix} \end{equation*}

SolutionDetermine if the linear transformation \(\ltdefn{S}{P_3}{M_{22}}\) is (a) injective, (b) surjective, (c) invertible. \begin{equation*} \lteval{S}{a+bx+cx^2+dx^3}= \begin{bmatrix} -a+4b+c+2d & 4a-b+6c-d\\ a+5b-2c+2d & a+2c+5d \end{bmatrix} \end{equation*}

SolutionFor each linear transformation below: (a) Find the matrix representation of \(T\text{,}\) (b) Calculate \(n(T)\text{,}\) (c) Calculate \(r(T)\text{,}\) (d) Graph the image in either \(\real{2}\) or \(\real{3}\) as appropriate, (e) How many dimensions are lost?, and (f) How many dimensions are preserved?

- \(\ltdefn{T}{\complex{3}}{\complex{3}}\) given by \(\lteval{T}{\colvector{x\\y\\z}} = \colvector{x\\x\\x}\)
- \(\ltdefn{T}{\complex{3}}{\complex{3}}\) given by \(\lteval{T}{\colvector{x\\y\\z}} = \colvector{x\\y\\0}\)
- \(\ltdefn{T}{\complex{3}}{\complex{2}}\) given by \(\lteval{T}{\colvector{x\\y\\z}} = \colvector{x\\x}\)
- \(\ltdefn{T}{\complex{3}}{\complex{2}}\) given by \(\lteval{T}{\colvector{x\\y\\z}} = \colvector{x\\y}\)
- \(\ltdefn{T}{\complex{2}}{\complex{3}}\) given by \(\lteval{T}{\colvector{x\\y}} = \colvector{x\\y\\0}\)
- \(\ltdefn{T}{\complex{2}}{\complex{3}}\) given by \(\lteval{T}{\colvector{x\\y}} = \colvector{x\\y\\x+y}\)

Consider the linear transformation \(\ltdefn{S}{M_{12}}{P_1}\) from the set of \(1\times 2\) matrices to the set of polynomials of degree at most 1, defined by \begin{equation*} \lteval{S}{\begin{bmatrix}a & b\end{bmatrix}}=(3a+b)+(5a+2b)x\text{.} \end{equation*} Prove that \(S\) is invertible. Then show that the linear transformation \begin{equation*} \ltdefn{R}{P_1}{M_{12}},\quad \lteval{R}{r+sx}=\begin{bmatrix}(2r-s) & (-5r+3s)\end{bmatrix} \end{equation*} is the inverse of \(S\text{,}\) that is \(\ltinverse{S}=R\text{.}\)

SolutionThe linear transformation \(S\) below is invertible. Find a formula for the inverse linear transformation, \(\ltinverse{S}\text{.}\) \begin{equation*} \ltdefn{S}{P_1}{M_{12}},\quad \lteval{S}{a+bx}=\begin{bmatrix} 3a+b & 2a+b \end{bmatrix}\text{.} \end{equation*}

SolutionThe linear transformation \(\ltdefn{R}{M_{12}}{M_{21}}\) is invertible. Determine a formula for the inverse linear transformation \(\ltdefn{\ltinverse{R}}{M_{21}}{M_{12}}\text{.}\) \begin{equation*} \lteval{R}{\begin{bmatrix}a & b\end{bmatrix}} = \begin{bmatrix} a+3b\\ 4a+11b \end{bmatrix} \end{equation*}

SolutionRework Example CIVLT, only in place of the basis \(B\) for \(P_2\text{,}\) choose instead to use the basis \(C=\set{1,\,1+x,\,1+x+x^2}\text{.}\) This will complicate writing a generic element of the domain of \(\ltinverse{T}\) as a linear combination of the basis elements, and the algebra will be a bit messier, but in the end you should obtain the same formula for \(\ltinverse{T}\text{.}\) The inverse linear transformation is what it is, and the choice of a particular basis should not influence the outcome.

Suppose \(U\) and \(V\) are vector spaces. Define the function \(\ltdefn{Z}{U}{V}\) by \(\lteval{Z}{\vect{u}}=\zerovector_{V}\) for every \(\vect{u}\in U\text{.}\) Then by Exercise LT.M60, \(Z\) is a linear transformation. Formulate a condition on \(U\) and \(V\) that is equivalent to \(Z\) being an invertible linear transformation. In other words, fill in the blank to complete the following statement (and then give a proof): \(Z\) is invertible if and only if \(U\) and \(V\) are . (See Exercise ILT.M60, Exercise SLT.M60, Exercise MR.M60.)

Prove that the identity linear transformation (Definition IDLT) is both injective and surjective, and hence invertible.

Suppose that \(\ltdefn{T}{U}{V}\) is a surjective linear transformation and \(\dimension{U}=\dimension{V}\text{.}\) Prove that \(T\) is injective.

SolutionSuppose that \(\ltdefn{T}{U}{V}\) is an injective linear transformation and \(\dimension{U}=\dimension{V}\text{.}\) Prove that \(T\) is surjective.

Suppose that \(U\) and \(V\) are isomorphic vector spaces, not of dimension zero. Prove that there are infinitely many isomorphisms between \(U\) and \(V\text{.}\)

SolutionSuppose \(\ltdefn{T}{U}{V}\) and \(\ltdefn{S}{V}{W}\) are linear transformations and \(\dimension{U}=\dimension{V}=\dimension{W}\text{.}\) Suppose that \(\compose{S}{T}\) is invertible. Prove that \(S\) and \(T\) are individually invertible (this could be construed as a converse of Theorem CIVLT).

Solution