Having introduced linear transformations, we now treat them as proper objects of study. Forget for a moment the linear algebraic nature of a linear transformation \(T\colon V\rightarrow W\text{,}\) and think of it just as a function. Purely along function-theoretic lines, we want to know whether \(T\) is injective, surjective, and invertible. As we will learn, there are two subspaces associated to a linear transformation \(T\text{,}\) its null space and image, that provide an easy way for answering these questions. We will also see that in the case of a matrix transformation \(T_A\text{,}\) these associated spaces coincide with two of the fundamental spaces of the matrix \(A\text{.}\) (You can probably guess one of these.)
As with the fundamental spaces of a matrix, given a linear transformation \(T\colon V\rightarrow W\) it is helpful to keep straight the different ambient spaces where \(\NS T\) and \(\im T\) live. As illustrated by Figure 5.2.(a), we have \(\NS T\subseteq V\) and \(\im T\subseteq W\text{:}\) that is, the null space is a subset of the domain of \(T\text{,}\) and the image is a subset of the codomain. Figures 5.2.(b)and 5.2.(c) go on to convey that \(\NS T\) is the set of elements of \(V\) that are mapped to \(\boldzero_W\text{,}\) and that \(\im T\) is the set of outputs of \(T\text{.}\)
Since \(T(\boldzero_V)=\boldzero_W\) (Theorem 3.2.12), we see that \(\boldzero_W\) is “hit” by \(T\text{,}\) and hence is a member of \(\im T\text{.}\)
Assume vectors \(\boldw_1, \boldw_2\in W\) are elements of \(\im T\text{.}\) By definition, this means there are vectors \(\boldv_1, \boldv_2\in V\) such that \(T(\boldv_i)=\boldw_i\) for \(1\leq i\leq 2\text{.}\) Now given any linear combination \(\boldw=c\boldw_1+d\boldw_2\text{,}\) we have
This shows that for any linear combination \(\boldw=c\boldw_1+d\boldw_2\text{,}\) there is an element \(\boldv=c\boldv_1+d\boldv_2\) such that \(T(\boldv)=\boldw\text{.}\) We conclude that if \(\boldw_1, \boldw_2\in \im T\text{,}\) then \(\boldw=c\boldw_1+d\boldw_2\in \im T\) for any \(c,d\in \R\text{,}\) as desired.
Let \(W=\{B\in M_{nn}\colon B^T=-B\}\text{,}\) subspace of skew-symmetric \(n\times n\) matrices. We claim \(\im F=W\text{.}\) As this is a set equality, we prove it by showing the two set inclusions \(\im F\subseteq W\) and \(W\subseteq \im F\text{.}\) (See Basic set properties)
The inclusion \(\im F\subseteq W\) is the easier of the two. If \(B\in \im F\text{,}\) then \(B=F(A)=A^T-A\) for some \(A\in M_{nn}\text{.}\) Using various properties of transposition, we have
The inclusion \(W\subseteq \im F\) is trickier: we must show that if \(B\) is skew-symmetric, then there is an \(A\) such that \(B=F(A)=A^T-A\text{.}\) Assume we have a \(B\) with \(B^T=-B\text{.}\) Letting \(A=-\frac{1}{2}B\) we have
As illustrated by Example 5.2.4, Theorem 5.2.3 provides an alternative technique for proving that a subset of \(W\subseteq V\) is in fact a subspace: namely, find a linear transformation \(T\colon V\rightarrow Z\) such that \(W=\NS T\text{.}\)
Not surprisingly, there is a connection between the null space of a matrix, as defined in Definition 4.1.10, and our new notion of null space. Indeed, given an \(m\times n\) matrix \(A\text{,}\) for all \(\boldx\in \R^n\) we have
and let \(T_A\colon \R^4\rightarrow \R^2\) be its associated matrix transformation. Provide bases for \(\NS T_A\) and \(\im T_A\) and compute the dimensions of these spaces.
The rank-nullity theorem relates the the dimensions of the null space and image of a linear transformation \(T\colon V\rightarrow W\text{,}\) assuming \(V\) is finite dimensional. Roughly speaking, it says that the bigger the null space, the smaller the image. More precisely, it tells us that
As we will see, this elegant result can be used to significantly simplify computations with linear transformations. For example, in a situation where we wish to compute explicitly both the null space and image of a given linear transformation, we can often get away with just computing one of the two spaces and using the rank-nullity theorem (and a dimension argument) to easily determine the other. Additionally, the rank-nullity theorem will directly imply some intuitively obvious properties of linear transformations. For example, suppose \(V\) is a finite-dimensional vector space. It seems obvious that if \(\dim W> \dim V\text{,}\) then there is no linear transformation mapping \(V\) surjectively onto \(W\text{:}\) i.e., you should not be able to map a “smaller” vector space onto a “bigger” one. Similarly, if \(\dim W \lt \dim V\text{,}\) then we expect that there is no injective linear transformation mapping \(V\) injectively into \(W\text{.}\) Both these results are easy consequences of the rank-nullity theorem.
Choose a basis \(B'=\{\boldv_1, \boldv_2, \dots, \boldv_k\}\) of \(\NS T\) and extend \(B'\) to a basis \(B=\{\boldv_1, \boldv_2,\dots, \boldv_k,\boldv_{k+1},\dots, \boldv_n\}\text{,}\) using Theorem 4.4.15. Observe that \(\dim\NS T=\nullity T=k\) and \(\dim V=n\text{.}\)
Suppose \(a_kT(\boldv_k)+a_{k+1}T(\boldv_{k+1})+\cdots +a_nT(\boldv_n)=\boldzero\text{.}\) Then the vector \(\boldv=a_k\boldv_k+a_{k+1}\boldv_{k+1}+\cdots +a_n\boldv_n\) satisfies \(T(\boldv)=\boldzero\) (using linearity of \(T\)), and hence \(\boldv\in \NS T\text{.}\) Then, using the fact that \(B'\) is a basis of \(\NS T\text{,}\) we have
Since the set \(B\) is linearly independent, we conclude that \(b_i=a_j=0\) for all \(1\leq i\leq k\) and \(k+1\leq j\leq n\text{.}\) In particular, \(a_{k+1}=a_{k+2}=\cdots=a_n=0\text{,}\) as desired.
It is clear that \(\Span B''\subseteq \im T\) since \(T(\boldv_i)\in \im T\) for all \(k+1\leq i\leq n\) and \(\im T\) is closed under linear combinations.
For the other direction, suppose \(\boldw\in \im T\text{.}\) Then there is a \(\boldv\in V\) such that \(\boldw=T(\boldv)\text{.}\) Since \(B\) is a basis of \(V\) we may write
Since \(\im T\subseteq \R^3\) and \(\dim\im T=\dim \R^3=3\text{,}\) we conclude by Corollary 4.4.17 that \(\im T=\R^3\text{.}\) Thus \(T\) is surjective.
Let \(F\colon M_{nn}\rightarrow M_{nn}\) be defined as \(F(A)=A^T-A\) as in Example 5.2.4. Prove that \(\im F\) is the subspace \(W\) of all skew-symmetric matrices following the steps below.
Thus \(\NS F\) is the subspace of all symmetric matrices. It can be shown that this space has dimension \(\frac{n(n+1)}{2}\text{.}\) To see why this is true intuitively, note that a symmetric matrix is completely determined by the entries on and above the diagonal: there are
As we showed in Example 5.2.4, any matrix \(B=F(A)\in \im F\) satisfies \(B^T=-B\text{:}\) i.e., \(\im F\subseteq W\text{,}\) where \(W\) is the subspace of all skew-symmetric \(n\times n\) matrices. Similarly to the space of symmetric matrices, we can show that \(\dim W=\frac{n(n-1)}{2}\text{:}\) intuitively, a skew-symmetric matrix is determined by the \(n(n-1)/2\) entries strictly above the diagonal (the diagonal entries must be equal to zero). Since \(\im F\subseteq W\) and \(\dim \im F=\dim W=\frac{n(n-1)}{2}\text{,}\) we conclude that \(\im F=W\text{,}\) by Corollary 4.4.17. This proves that \(\im F\) is the space of all skew-symmetric matrices.
Subsection5.2.3Injective and surjective linear transformations
Recall the notions of injectivity and surjectivity from Definition 0.2.7: a function \(f\colon X\rightarrow Y\) is injective (or one-to-one) if for all \(x,x'\in X\) we have \(f(x)=f(x')\) implies \(x=x'\text{;}\) it is surjective (or onto) if for all \(y\in Y\) there is an \(x\in X\) with \(f(x)=y\text{.}\) As with all functions, we will be interested to know whether a given linear transformation is injective or surjective; as it turns out, the concepts of null space and image give us a convenient manner of answering these questions. As remarked in Definition 0.2.7, there is in general a direct connection between the surjectivity and the image of a function: namely, \(f\colon X\rightarrow Y\) is surjective if and only if \(\im f=Y\text{.}\) It follows immediately that a linear transformation \(T\colon V\rightarrow W\) is surjective if and only if \(\im T=W\text{.}\) As for injectivity, it is easy to see that if a linear transformation \(T\) is injective, then its null space must consist of just the zero vector of \(V\text{.}\) What is somewhat surprising is that the converse is also true, as we now show.
Statement (2) is true of any function, whether it is a linear transformation or not; it follows directly from the definitions of surjectivity and image. Thus it remains to prove statement (1). We prove both implications separately.
To determine whether a function of sets \(f\colon X\rightarrow Y\) is injective, we normally have to show that for each output \(y\) in the image of \(f\) there is exactly one input \(x\) satisfying \(f(x)=y\text{.}\) Think of this as checking injectivity at every output. Theorem 5.2.12 tells us that in the special case of a linear transformation \(T\colon V\rightarrow W\) it is enough to check injectivity at exactly one ouput: namely, \(\boldzero\in W\text{.}\)
Let \(V\) and \(W\) be vector spaces. A linear transformation \(T\colon V\rightarrow W\) is an isomorphism if it is invertible as a function: i.e., if there is an inverse function \(T^{-1}\colon W\rightarrow V\) satisfying
Definition 5.2.15 leaves open the question of whether the inverse function \(T^{-1}\colon W\rightarrow V\) of an isomorphism \(T\colon V\rightarrow W\) is a linear transformation. The next theorem resolves that issue.
Let \(T\colon V\rightarrow W\) be an isomorphism, and let \(T^{-1}\colon W\rightarrow V\) be its inverse function. We use the 1-step technique to show \(T^{-1}\) is linear, but with a slight twist: given any \(c,d\in \R\) and vectors \(\boldw_1,\boldw_2\in W\) we will show that
According to Definition 5.2.15, to prove a function \(T\colon V\rightarrow W\) is an isomorphism, we must show both that \(T\) is linear and that it is invertible. We know already how to decide whether a function is linear, but how do we decide whether it is invertible? Recall that a function is invertible if and only if it is bijective. (See Theorem 0.2.11.) This fact gives rise to two distinct methods for proving that a given linear transformation is invertible:
we can show directly that \(T\) is invertible by providing an inverse \(T^{-1}\colon W\rightarrow V\text{;}\)
One of the two methods described in Remark 5.2.17 for determining whether a linear transformation is invertible may be more convenient than the other, depending on the linear transformation in question. Thanks to Theorem 5.2.12, we can decide whether a linear transformation is bijective by looking at its two associated subspaces \(\NS T\) and \(\im T\text{.}\)
The result follows directly from Theorem 5.2.12 and the fact that a function is invertible if and only if it is bijective:
\begin{align*}
T \text{ is an isomorphism} \amp \iff T \text{ is bijective}\\
\amp \iff T \text{ is injective and surjective}\\
\amp \iff \NS T=\{\boldzero_V\} \text{ and } \im T=W\text{.}
\end{align*}
We know already that \(F\) is a linear transformation. It remains to show that it is invertible. In the spirit of Remark 5.2.17, we give two proofs, corresponding to methods (1) and (2).
Define \(G\colon M_{nm}\rightarrow M_{mn}\) as \(G(B)=B^T\text{.}\) We claim that \(G=F^{-1}\text{,}\) and hence that \(F\) is an isomorphism. To do show we must show
Why is it useful to know whether two vector spaces are isomorphic? The short answer is that if \(V\) and \(W\) are isomorphic, then although they may be very different objects when considered as sets, from the linear-algebraic perspective there is essentially no difference between the two: i.e., they satisfy the exact same linear-algebraic properties. Furthermore, an isomorphism \(T\colon V\rightarrow W\) witnessing the fact that \(V\) and \(W\) are isomorphic gives us a perfect bijective dictionary between the two spaces, allowing us to answer questions about the one space by “translating” it to a question about the other, using \(T\) or \(T^{-1}\text{.}\)Theorem 5.2.20 gives a first glimpse into this dictionary-like nature of isomorphisms. Later in Section 5.3 we will introduce a particular isomorphism called the coordinate vector, which illustrates the computational value of being able to translate questions about abstract vector spaces \(V\) to questions about our beloved and familiar \(\R^n\) spaces.
We prove \((1)\) and leave the remaining statements as an exercise. Let \(T\colon V\rightarrow W\) be an isomorphism, and let \(S\subseteq V \) be a linearly independent set. Define
Let \(T^{-1}\colon W\rightarrow V\) be the inverse function of \(T\text{,}\) and recall that \(T^{-1}\) is linear by Theorem 5.2.16. Now applying \(T^{-1}\) to both sides of the equation above and simplify
Assume \(W\) is a vector space satisfying \(\dim V=\dim W=n\text{,}\) and let \(T\colon V\rightarrow W\) be a linear transformation. The following are equivalent.
If \(V\) and \(W\) are isomorphic, then there is an isomorphism \(T\colon V\rightarrow W\text{.}\) It follows from Theorem 5.2.20 that \(\dim V=\dim W\text{.}\)
Now assume \(\dim V=\dim W=n\text{.}\) By definition there are bases \(B=\{\boldv_1,\boldv_2,\dots, \boldv_n\}\) and \(B'=\{\boldw_1,\boldw_2,\dots, \boldw_n\}\) of \(V\) and \(W\text{,}\) respectively, satisfying \(\abs{B}=\abs{B'}=n\text{.}\) Using Existence of transformations there is a linear transformation
\begin{align*}
T\colon V \amp \rightarrow W
\end{align*}
satisfying \(T(\boldv_i)=\boldw_i\) for all \(1\leq i\leq n\text{,}\) and a linear transformation
\begin{align*}
T'\colon W\rightarrow \amp V
\end{align*}
satisfying \(T'(\boldw_i)=\boldv_i\) for all \(1\leq i\leq n\text{.}\) We now use Uniqueness of transformations to prove that
\begin{align*}
T'\circ T \amp =\id_V \amp T\circ T'\amp =\id_W\text{,}
\end{align*}
and hence that \(T'=T^{-1}\text{.}\) To do so we observe that for all \(1\leq i\leq n\) we have
\begin{align*}
T'\circ T(\boldv_i) \amp = T'(\boldw_i) \amp (\text{def. of } T)\\
\amp = \boldv_i \amp (\text{def. of } T') \\
\amp = \id_V(\boldv_i) \amp (\text{def. of } \id_V\text{.}
\end{align*}
Since \(T'\circ T\) and \(\id_V\) agree on the basis \(B\) (\(T'\circ T(\boldv_i)=\id_V(\boldv_i)\) for all \(1\leq i\leq n\)), we conclude that \(T'\circ T=\id V\text{.}\) A very similar argument shows \(T\circ T'=\id_W\text{.}\) Having shown that \(T'=T^{-1}\text{,}\) we conclude that \(T\) is an isomorphism, and hence that \(V\) and \(W\) are isomorphic.
Assume that \(T\colon V\rightarrow W\) and that \(\dim V=\dim W=n\text{.}\) We will establish the cycle of implications (a)\(\implies\)(b)\(\implies\)(c)\(\implies\)(a).
A shocking consequence of Theorem 5.2.22, is that any two vector spaces of the same dimension are isomorphic: in particular, any \(n\)-dimensional vector space is isomorphic to \(\R^n\text{!}\)
We have finally delivered on a promise made way back in Chapter 1: namely, we now see how any finite-dimensional vector space is isomorphic (and thus structurally equivalent) to one of our Euclidean spaces \(\R^n\text{.}\) There is something almost anticlimactic about Corollary 5.2.23. Having devoted much time and energy to introducing and studying various “exotic” finite-dimensional vector spaces, we now learn that they are essentially no different than our familiar spaces \(\R^n\text{.}\) Be not disappointed! By establishing an isomorphism between an exotic vector space \(V\) and a Euclidean space \(\R^n\text{,}\) we are able to transport to \(V\) all the wonderful computational techniques we have at our disposal in the \(\R^n\) context. Furthermore, it is still up to us to choose our isomorphism \(T\colon V\rightarrow \R^n\text{,}\) and as we will see, there is a subtle art to choosing the isomorphism \(T\) to suit our particular needs.
We end this section by making a connection between invertible matrices and isomorphisms. It follows easily from Theorem 5.2.22 that a matrix \(A\in M_{nn}\) is invertible if and only if its corresponding matrix transformation \(T_A\colon \R^n\rightarrow \R^n\) is an isomorphism. Indeed, \(T_A\) is an isomorphism if and only if \(\NS T_A=\{\boldzero\}\text{,}\) if and only if \(\NS A=\{\boldzero\}\text{,}\) if and only if \(A\) is invertible. Furthermore, we see in this case that the inverse function of \(T_A\) is the matrix transformation \(T_{A^{-1}}\text{:}\) i.e., \((T_A)^{-1}=T_{A^{-1}}\text{.}\) We add this statement to our ever-growing list of equivalent formulations of invertibility.
Any of the following equivalent conditions about the set \(S\) of columns of \(A\) hold: \(S\) is a basis of \(\R^n\text{;}\)\(S\) spans \(\R^n\text{;}\)\(S\) is linearly independent.
Any of the following equivalent conditions about the set \(S\) of rows of \(A\) hold: \(S\) is a basis of \(\R^n\text{;}\)\(S\) spans \(\R^n\text{;}\)\(S\) is linearly independent.
If \(T:{\mathbb R}^8\to {\mathbb R}^2\) is a linear transformation, then consider whether the set ker (\(T\) ) is a subspace of \({\mathbb R}^{8}\text{.}\)
Define the linear transformation \(T: {\mathbb R}^2 \rightarrow {\mathbb R}^3\) by \(T(\vec{x}) = A\vec{x}\text{.}\) Find a vector \(\vec{x}\) whose image under \(T\) is \(\vec{b}\text{.}\)
For each linear transformation \(T\) give parametric descriptions of \(\NS T\) and \(\im T\text{.}\) To do so you will want to relate each computation to a system of linear equations.
For each subset \(W\subseteq V\) show \(W\) is a subspace by identifying it with the null space of a linear transformation \(T\text{.}\) You may use any of the examples from Section 5.1, and any of the results from the exercises in Exercises 5.1.6.
