6 Linear maps

6.1 Linear maps

Throughout this section, $V,W$ denote $\mathbb{K}$-vector spaces. So we have a notion of addition and scalar multiplication; and if we have a mapping $f : V \to W,$ we can ask if it respects these structures.

Definition 6.1 • Linear map

A mapping $f : V \to W$ is called linear if it satisfies the following two conditions:

it is additive, i.e. for all $v_1, v_2 \in V$ we have \[f(v_1 + v_2) = f(v_1) + f(v_2).\]
It is $1$-homogeneous, that is, for all $s \in \mathbb{K}$ and $v \in V$ we have \[f(s v) = s f(v).\]

One can check (see Exercises) that a mapping $f : V \to W$ is linear iff it satisfies \[\tag{6.1} f(s_1v_1+s_2v_2)=s_1 f(v_1)+s_2 f(v_2)\] for all $s_1, s_2 \in \mathbb{K}$ and $v_1, v_2 \in V.$

Example 6.2

Notice that “most” functions $\mathbb{R}\to \mathbb{R}$ are neither additive nor $1$-homogeneous. As an example, consider a mapping $f : \mathbb{R}\to \mathbb{R}$ which satisfies the $1$-homogeneity property. Let $a=f(1) \in \mathbb{R}.$ Then the $1$-homogeneity implies that for all $x \in \mathbb{R}=\mathbb{R}^1$ we have \[f(x)=f(x\cdot 1)=x\cdot f(1)=a \cdot x,\] showing that the only $1$-homogeneous mappings from $\mathbb{R}\to \mathbb{R}$ are of the form $x \mapsto a x,$ where $a$ is a real number. In particular, $\sin,\cos,\tan,\log,\exp,\sqrt{\phantom{x}}$ and all polynomials of degree higher than one are not linear.

Examples

For instance, you already saw in Algorithmics that matrices give linear maps: if $\mathbf{M}\in M_{m,n}(\mathbb{K}),$ then for any $\vec{x}, \vec{y} \in \mathbb{K}^n$ and $a, b \in \mathbb{K},$ we have \[\mathbf{M}\cdot (a \vec{x} + b \vec{y}) = a \mathbf{M}\vec{x} + b \mathbf{M}\vec{y},\] so the map $f_{\mathbf{M}} : \mathbb{K}^n \to \mathbb{K}^m$ defined by $f_{\mathbf{M}}(\vec{x}) = \mathbf{M}\vec{x}$ is linear. However, working with abstract vector spaces means we have all kinds of other exciting linear maps to study.

Example 6.3

If $\mathsf{P}(\mathbb{R})$ is the vector space of polynomials from 4.1, then we may think of the derivative with respect to the variable $x$ as a mapping \[\frac{\mathrm{d}}{\mathrm{d}x} : \mathsf{P}(\mathbb{R}) \to \mathsf{P}(\mathbb{R}).\] Now recall that the derivative satisfies \[\tag{6.2} \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}(p+q)&=\frac{\mathrm{d}}{\mathrm{d}x}(p)+\frac{\mathrm{d}}{\mathrm{d}x}(q) \qquad && (\text{additivity}),\\ \frac{\mathrm{d}}{\mathrm{d}x}(s\cdot p)&=s\cdot \frac{\mathrm{d}}{\mathrm{d}x}(p)\qquad && (\text{$1$-homogeneity}). \end{aligned}\] so it is indeed linear.

Example 6.4

The matrix transpose is a map $M_{m,n}(\mathbb{K}) \to M_{n,m}(\mathbb{K})$ and this map is linear. Indeed, for all $s,t \in \mathbb{K}$ and $\mathbf{A},\mathbf{B}\in M_{m,n}(\mathbb{K}),$ we have \[\begin{gathered} (s \mathbf{A}+t \mathbf{B})^T=(s A_{ji}+t B_{ji})_{1\leqslant j\leqslant n, 1\leqslant i \leqslant m}=s(A_{ji})_{1\leqslant j\leqslant n, 1\leqslant i \leqslant m}+\\t(B_{ji})_{1\leqslant j\leqslant n, 1\leqslant i \leqslant m}=s \mathbf{A}^T+t \mathbf{B}^T. \end{gathered}\]

Example 6.5

Let $V$ be a $\mathbb{K}$-vector space. Then the identity mapping $\mathrm{Id}_V : V \to V$ is linear, since for all $s_1,s_2 \in \mathbb{K}$ and $v_1,v_2 \in V$ we have \[\mathrm{Id}_V(s_1v_1+s_2v_2)=s_1v_1+s_2v_2 = s_1\mathrm{Id}_V(v_1)+s_2\mathrm{Id}_V(v_2).\]

First properties of linear maps

A necessary condition for linearity of a mapping is that it maps the zero vector onto the zero vector:

Lemma 6.6

Let $f : V \to W$ be a linear map, then $f(0_V)=0_W.$

Proof. Since $f : V \to W$ is linear, we have \[f(0_V)=f(0\cdot 0_V)=0\cdot f(0_V)=0_W. \]

We’ll now investigate what happens if you compose two such maps. Recall that if $f : \mathcal{X}\to \mathcal{Y}$ is a mapping from a set $\mathcal{X}$ into a set $\mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ a mapping from $\mathcal{Y}$ into a set $\mathcal{Z},$ we can consider the composition of $g$ and $f$ \[g\circ f : \mathcal{X} \to \mathcal{Z}, \qquad x \mapsto g(f(x)).\] Recall also that a mapping is bijective if and only if it has an inverse mapping, which is a map $f^{-1} : \mathcal{Y}\to \mathcal{X}$ with $f^{-1} \circ f = \mathrm{Id}_\mathcal{X}$ and $f \circ f^{-1} = \mathrm{Id}_\mathcal{Y}$ (and this inverse mapping is unique if it exists).

Proposition 6.7

Let $V_1,V_2,V_3$ be $\mathbb{K}$-vector spaces and $f : V_1 \to V_2$ and $g: V_2 \to V_3$ be linear maps. Then the composition $g \circ f : V_1 \to V_3$ is linear. Furthermore, if $f : V_1 \to V_2$ is bijective, then the inverse map $f^{-1} : V_2 \to V_1$ is linear.

Proof. For the first statement, let $s,t \in \mathbb{K}$ and $v,w \in V_1.$ Then \[\begin{aligned} \left(g\circ f\right)(s v+t w)&=g(f(s v+t w))=g(s f(v)+t f(w))\\&=s g(f(v))+t g(f(w))=s(g\circ f)(v)+t(g\circ f)(w), \end{aligned}\] where we first use the linearity of $f$ and then the linearity of $g.$ It follows that $g\circ f$ is linear.

The second statement is more delicate. Recall that for any $v \in V_2$ we have $f (f^{-1}(v)) = v.$ So for $v, w \in V_2$ we can write \[f (f^{-1}(s v + t w)) = sv + tw = s f(f^{-1}(v)) + t f (f^{-1}(w)).\] Using linearity of $f,$ the right-hand side can be rewritten as \[s f(f^{-1}(v)) + t f (f^{-1}(w)) = f(s f^{-1}(v) + t f^{-1}(w)).\] Thus \[f (f^{-1}(s v + t w)) = f(s f^{-1}(v) + t f^{-1}(w))\] and since $f$ is injective, we can cancel the $f$’s to conclude \[f^{-1}(s v + t w) = s f^{-1}(v) + t f^{-1}(w).\]

Isomorphisms

Bijective linear maps are particularly important and they have a special name:

Definition 6.8 • Vector space isomorphism

A bijective linear map $f : V \to W$ is called a (vector space) isomorphism. If an isomorphism $f : V \to W$ exists, then the $\mathbb{K}$-vector spaces $V$ and $W$ are called isomorphic.

Remark 6.9

If two vector spaces are isomorphic, then they are “the same as vector spaces”. For instance, the obvious map $\mathbb{K}_n \to \mathbb{K}^n$ given by $(x_1\ \dots \ x_n) \mapsto \left(\begin{smallmatrix} x_1 \\ \vdots \\ x_n\end{smallmatrix}\right)$ is an isomorphism.

6.2 Images, preimages, kernels

In this section we’ll investigate how vector subspaces interact with linear maps.

Proposition 6.10

Let $V,W$ be $\mathbb{K}$-vector spaces, $U\subset V$ and $Z\subset W$ be vector subspaces and $f : V \to W$ a linear map. Then the image $f(U)$ is a vector subspace of $W$ and the preimage $f^{-1}(Z)$ is a vector subspace of $V.$

Proof. Since $U$ is a vector subspace, we have $0_V \in U.$ By Lemma 6.6, $f(0_V)=0_W,$ hence $0_W \in f(U).$ For all $w_1,w_2 \in f(U)$ there exist $u_1,u_2 \in U$ with $f(u_1)=w_1$ and $f(u_2)=w_2.$ Hence for all $s_1,s_2 \in \mathbb{K}$ we obtain \[s_1w_1+s_2w_2=s_1f(u_1)+s_2f(u_2)=f(s_1u_1+s_2u_2),\] where we use the linearity of $f.$ Since $U$ is a subspace, $s_1u_1+s_2u_2$ is an element of $U$ as well. It follows that $s_1w_1+s_2 w_2 \in f(U)$ and hence applying Definition 4.16 again, we conclude that $f(U)$ is a subspace of $W.$ The second claim is left to the reader as an exercise.

We now define some special subspaces associated to a linear map.

Definition 6.11 • Kernel

The kernel of a linear map $f : V \to W$ is the preimage of $\{0_W\}$ under $f,$ that is, \[\operatorname{Ker}(f)=\left\{v \in V\,|\, f(v)=0_W\right\}=f^{-1}(\{0_W\}).\]

Example 6.12

The kernel of the linear map $\frac{\mathrm{d}}{\mathrm{d}x} : \mathsf{P}_n(\mathbb{R}) \to \mathsf{P}_{n-1}(\mathbb{R})$ consists of the constant polynomials satisfying $f(x)=c$ for all $x \in \mathbb{R}$ and where $c \in \mathbb{R}$ is some constant.

An immediate consequence of Proposition 6.10 is:

Corollary 6.13

Let $f : V \to W$ be a linear map, then its image $\operatorname{Im}(f)$ is a vector subspace of $W$ and its kernel $\operatorname{Ker}(f)$ is a vector subspace of $V.$

Roughly, you can think of the kernel as being the stuff which gets “lost in the machine”, and the image as the stuff which “comes out the other side”. So these should, in some sense, add up to all of $V$; and we’ll make this precise a little later (in Theorem 6.20).

We can characterise the injectivity of a linear map $f : V \to W$ in terms of its kernel:

Lemma 6.14

A linear map $f : V \to W$ is injective if and only if $\operatorname{Ker}(f)=\{0_V\}.$

Proof. Let $f : V \to W$ be injective. Suppose $f(v)=0_W.$ Since $f(0_V)=0_W$ by Lemma 6.6, we have $f(v)=f(0_V),$ hence $v=0_V$ by the injectivity assumption. It follows that $\operatorname{Ker}(f)=\{0_V\}.$ Conversely, suppose $\operatorname{Ker}(f)=\{0_V\}$ and let $v_1,v_2 \in V$ be such that $f(v_1)=f(v_2).$ Then by the linearity we have $f(v_1)-f(v_2)=0_W=f(v_1-v_2).$ Hence $v_1-v_2$ is in the kernel of $f$ so that $v_1-v_2=0_V$ or $v_1=v_2.$

We can characterise isomorphisms using kernels and images. By the definition of surjectivity, a map $f : V \to W$ is surjective if and only if $\operatorname{Im}(f)=W.$ Combining this with Lemma 6.14 gives:

Proposition 6.15

A linear map $f : V \to W$ is an isomorphism if and only if $\operatorname{Ker}(f)=\{0_V\}$ and $\operatorname{Im}(f)=W.$

Linear maps and dimension

Lemma 6.16

Let $f : V\to W$ be linear.

If $\mathcal{S}\subset V$ is a generating set, and $f$ is surjective, then $f(\mathcal{S})$ is a generating set of $W.$
If $f$ is surjective, and $V$ is finite-dimensional, then $W$ is finite-dimensional.
If $\mathcal{S}\subset V$ is an LI set, and $f$ is injective, then $f(\mathcal{S})$ is an LI set in $W.$
If $f$ is injective, and $W$ is finite-dimensional, then $V$ is finite-dimensional.

Proof. (i) Let $w \in W.$ Since $f$ is surjective there exists $v \in V$ such that $f(v)=w.$ Since $\operatorname{span}(\mathcal{S})=V,$ there exists $k \in \mathbb{N},$ as well as elements $v_1,\ldots,v_k \in \mathcal{S}$ and scalars $s_1,\ldots,s_k$ such that $v=\sum_{i=1}^ks_iv_i$ and hence $w=\sum_{i=1}^ks_if(v_i),$ where we use the linearity of $f.$ We conclude that $w \in \operatorname{span}(f(\mathcal{S}))$ and since $w$ is arbitrary, it follows that $W=\operatorname{span}(f(\mathcal{S})).$

(ii) If $V$ is finite-dimensional then it has a finite generating set $\mathcal{S}.$ Then $f(\mathcal{S})$ is a finite set in $W,$ and by (i) it is a generating set.

(iii) Suppose, for contradiction, that $f(\mathcal{S})$ is linearly dependent. Then we can find scalars $s_1, \dots, s_k$ (not all zero) and elements $w_1, \dots, w_s$ in $f(\mathcal{S})$ with $\sum s_i w_i = 0_W.$ But each $w_i$ must be $f(v_i)$ for some $v_i \in \mathcal{S},$ and $0_W = f(0_V),$ so we have $f(\sum s_i v_i) = \sum s_i f(v_i) = 0_W = f(0_V).$ Since $f$ is injective, we can conclude that $\sum s_i v_i = 0_V$ and thus $\mathcal{S}$ is itself linearly dependent, contradicting our assumption.

(iv) Suppose $V$ is infinite-dimensional. Then $V$ contains an infinite linearly independent set $\mathcal{S}$ by Exercise 5.1. By (iii), $f(\mathcal{S})$ is a linearly independent set in $W$; and it is still infinite, since $f$ is injective. So $W$ is infinite-dimensional.

Lemma 6.17

Let $f : V \to W$ be a vector space isomorphism. Then $V$ is finite-dimensional if and only if $W$ is; and if this holds, then $\dim(V) = \dim(W).$

Proof. Parts (ii) and (iv) of Lemma 6.16 show that $V$ is finite-dimensional iff $W$ is. Parts (i) and (iii) show that if $\mathcal{S}$ is a basis of $V,$ then $f(\mathcal{S})$ is a basis of $W$; but, since $f$ is injective, $f(\mathcal{S})$ has the same number of elements as $\mathcal{S}.$

Corollary 6.18

If $m \ne n,$ then $\mathbb{K}^m$ and $\mathbb{K}^n$ are not isomorphic as vector spaces.

Proof. We’ve seen that $\mathbb{K}^n$ has dimension $n,$ so if an isomorphism existed for $m \ne n$ it would contradict the previous lemma.

6.3 The rank-nullity theorem

Now that we know the notion of “dimension” is well-behaved, we’re going to study the dimensions of subspaces coming from a linear map.

Definition 6.19 • Rank and nullity for linear maps and matrices

Let $V,W$ be $\mathbb{K}$-vector spaces with $W$ finite dimensional. The rank and nullity of a linear map $f : V \to W$ are defined as \[\operatorname{rank}(f)=\dim \operatorname{Im}(f), \qquad \operatorname{nullity}(f) = \dim \operatorname{Ker}(f).\]

The following important theorem establishes a relation between the nullity and the rank of a linear map. It states something that is intuitively not surprising, namely that the dimension of the image of a linear map $f : V \to W$ is the dimension of the vector space $V$ minus the dimension of the subspace of vectors that we “lose”, that is, those that are mapped onto the zero vector of $W.$ More precisely:

Theorem 6.20 • Rank–nullity theorem

Let $V,W$ be finite dimensional $\mathbb{K}$-vector spaces and $f : V \to W$ a linear map. Then we have \[\dim(V)=\dim \operatorname{Ker}(f)+\dim \operatorname{Im}(f)=\operatorname{nullity}(f)+\operatorname{rank}(f).\]

Proof. Let $d=\dim \operatorname{Ker}(f)$ and $n=\dim V,$ so that $d \leqslant n$ by Proposition [prop:subspacedimension]. Let $\{v_1,\ldots,v_d\}$ be a basis of $\mathcal{S}=\operatorname{Ker}(f).$ By Theorem 5.10 (ii) we can find linearly independent vectors $\hat{\mathcal{S}}=\{v_{d+1},\ldots,v_n\}$ so that $\mathcal{T}=\mathcal{S}\cup \hat{\mathcal{S}}$ is a basis of $V.$ Now $U=\operatorname{span}(\hat{\mathcal{S}})$ is a subspace of $V$ of dimension $n-d.$ We consider the linear map \[g : U \to \operatorname{Im}(f), \quad v \mapsto f(v).\] We want to show that $g$ is an isomorphism, since then $\dim \operatorname{Im}(f)=\dim(U)=n-d,$ so that \[\dim \operatorname{Im}(f)=n-d=\dim(V)-\dim\operatorname{Ker}(f),\] as claimed.

We first show that $g$ is injective. Assume $g(v)=0_W.$ Since $v \in U,$ we can write $v=s_{d+1}v_{d+1}+\cdots +s_nv_n$ for scalars $s_{d+1},\ldots,s_n.$ Since $g(v)=0_W$ we have $v \in \operatorname{Ker}(f),$ hence we can also write $v=s_1v_1+\cdots +s_d v_d$ for scalars $s_1,\ldots,s_d,$ subtracting the two expressions for $v,$ we get \[0_V=s_1v_1+\cdots+s_d v_d-s_{d+1}v_{d+1}-\cdots-s_nv_n.\] Since $\{v_1,\ldots,v_n\}$ is a basis, it follows that all the coefficients $s_i$ vanish, where $1\leqslant i \leqslant n.$ Therefore we have $v=0_V$ and $g$ is injective.

Second, we show that $g$ is surjective. Suppose $w \in \operatorname{Im}(f)$ so that $w=f(v)$ for some vector $v \in V.$ We write $v=\sum_{i=1}^ns_iv_i$ for scalars $s_1,\ldots,s_n.$ Using the linearity of $f,$ we compute \[w=f(v)=f\left(\sum_{i=1}^ns_iv_i\right)=f\Big(\underbrace{\sum_{i=d+1}^ns_iv_i}_{=\hat{v}}\Big)=f(\hat{v})\] where $\hat{v} \in U.$ We thus have an element $\hat{v}$ with $g(\hat{v})=w.$ Since $w$ was arbitrary, we conclude that $g$ is surjective.

Corollary 6.21

Let $V,W$ be finite dimensional $\mathbb{K}$-vector spaces with $\dim(V)=\dim(W)$ and $f : V \to W$ a linear map. Then the following statements are equivalent:

$f$ is injective;
$f$ is surjective;
$f$ is bijective.

Proof. (i) $\Rightarrow$ (ii) By Lemma 6.14, the map $f$ is injective if and only if $\operatorname{Ker}(f)=\{0_V\}$ so that $\dim \operatorname{Ker}(f)=0$ by Example 5.14 (i). Theorem 6.20 implies that $\dim \operatorname{Im}(f)=\dim(V)=\dim(W)$ and hence Proposition [prop:subspacedimension] implies that $\operatorname{Im}(f)=W,$ that is, $f$ is surjective.

(ii) $\Rightarrow$ (iii) Since $f$ is surjective $\operatorname{Im}(f)=W$ and hence $\dim \operatorname{Im}(f)=\dim(W)=\dim(V).$ Theorem 6.20 implies that $\dim \operatorname{Ker}(f)=0$ so that $\operatorname{Ker}(f)=\{0_V\}$ by Proposition [prop:subspacedimension]. Applying Lemma 6.14 again shows that $f$ is injective and hence bijective.

(iii) $\Rightarrow$ (i) Since $f$ is bijective, it is also injective.

Corollary 6.22

Let $V,W$ be finite dimensional $\mathbb{K}$-vector spaces and $f : V \to W$ a linear map. Then $\operatorname{rank}(f)\leqslant \min\{\dim(V),\dim(W)\}$ and \[\begin{aligned} \operatorname{rank}(f)&=\dim(V)\iff f\text{ is injective,}\\ \operatorname{rank}(f)&=\dim(W)\iff f\text{ is surjective.} \end{aligned}\]

Proof. For the first claim it is sufficient to show that $\operatorname{rank}(f)\leqslant \dim(V)$ and $\operatorname{rank}(f)\leqslant \dim(W).$ By definition, $\operatorname{rank}(f)=\dim\operatorname{Im}(f)$ and since $\operatorname{Im}(f)\subset W,$ we have $\operatorname{rank}(f)=\dim\operatorname{Im}(f)\leqslant \dim(W)$ with equality if and only if $f$ is surjective, by Proposition [prop:subspacedimension].

Theorem 6.20 implies that $\operatorname{rank}(f)=\dim \operatorname{Im}(f)=\dim(V)-\dim \operatorname{Ker}(f)\leqslant \dim(V)$ with equality if and only if $\dim\operatorname{Ker}(f)=0,$ that is, when $f$ is injective (as we have just seen in the proof of the previous corollary).

Corollary 6.23

Let $V,W$ be finite dimensional $\mathbb{K}$-vector spaces and $f : V \to W$ a linear map. Then we have

If $\dim(V)<\dim(W),$ then $f$ is not surjective;
If $\dim(V)>\dim(W),$ then $f$ is not injective. In particular, there exist non-zero vectors $v \in V$ with $f(v)=0_W.$

Proof. (i) Suppose $\dim(V)<\dim(W),$ then by Theorem 6.20 \[\operatorname{rank}(f)=\dim(V)-\dim\operatorname{Ker}(f)\leqslant\dim(V)<\dim(W)\] and the claim follows from Corollary 6.22.

(ii) Suppose $\dim(V)>\dim(W),$ then \[\operatorname{rank}(f)\leqslant\dim(W)<\dim(V)\] and the claim follows from Corollary 6.22.

Proposition 6.24

Let $V,W$ be finite dimensional $\mathbb{K}$-vector spaces. Then there exists an isomorphism $\Theta : V \to W$ if and only if $\dim(V)=\dim(W).$

Proof. $\Rightarrow$ This was already proved in Lemma 6.17.

$\Leftarrow$ Let $\dim(V)=\dim(W)=n \in \mathbb{N}.$ Choose a basis $\mathcal{T}=\{w_1,\ldots,w_n\}$ of $W$ and consider the linear map \[\Theta : \mathbb{K}^n \to W, \quad \vec{x} \mapsto x_1w_1+\cdots+x_nw_n,\] where $\vec{x}=(x_i)_{1\leqslant i \leqslant n}$ Notice that $\Theta$ is injective. Indeed, if $\Theta(\vec{x})=x_1w_1+\cdots+x_nw_n=0_W,$ then $x_1=\cdots=x_n=0,$ since $\{w_1,\ldots,w_n\}$ are linearly independent. We thus conclude $\operatorname{Ker}\Theta=\{0_V\}$ and hence Lemma 6.14 implies that $\Theta$ is injective and therefore bijective by Corollary 6.21. The map $\Theta$ is linear and bijective, thus an isomorphism. Likewise, for a choice of basis $\mathcal{S}=\{v_1,\ldots,v_n\}$ of $V,$ we obtain an isomorphism $\Phi : \mathbb{K}^n \to V.$ Since the composition of bijective maps is again bijective, the map $\Theta\circ\Phi^{-1} : V \to W$ is bijective and since by Proposition 6.7 the composition of linear maps is again linear, the map $\Theta \circ \Phi^{-1} : V \to W$ is an isomorphism.

Exercises

Exercise 6.1

Let $f : V \to W$ be a linear map, $k\geqslant 2$ a natural number and $s_1,\ldots,s_k \in \mathbb{K}$ and $v_1,\ldots,v_k \in V.$ Show that $f : V \to W$ satisfies \[f(s_1v_1+\cdots +s_kv_k)=s_1f(v_1)+\cdots+s_kf(v_k)\] or written with the sum symbol \[\boxed{f\left(\sum_{i=1}^ks_iv_i\right)=\sum_{i=1}^ks_if(v_i).}\] This identity is used frequently in Linear Algebra, so make sure you understand it.

Solution

We will show the claim by induction on $k.$ If $k=2,$ the claim follows immediately by linearity of $f$ and the statement is anchored.

Inductive step: Suppose the statement is true for $k\geqslant 2$ and we will argue that it will be true for $k+1.$ Let $s_1,\ldots,s_{k+1}\in \mathbb{K}$ and $v_1,\ldots,v_{k+1}\in V.$ Writing $u = \sum_{i=1}^{k}s_iv_i,$ we have \[\begin{aligned} f\bigg(\sum_{i=1}^{k+1}s_iv_i\bigg) & = f(u+s_{k+1}v_{k+1}) \\ & = f(u) + s_{k+1}f(v_{k+1}) \\ & = \sum_{i=1}^{k}s_if(v_i) + s_{k+1}f(v_{k+1})\\ & = \sum_{i=1}^{k+1}s_if(v_i), \end{aligned}\] where the second equality follows by linearity of $f$ and the third one by the induction hypothesis.

Exercise 6.2

Show, conversely, that if a mapping $f : V \to W$ satisfies \[f(s_1v_1+s_2v_2)=s_1 f(v_1)+s_2 f(v_2)\] for all $s_1, s_2 \in \mathbb{K}$ and $v_1, v_2 \in V,$ then it is linear.

Solution

Setting $s_1 = s_2 = 1$ gives additivity, and setting $v_2 = 0$ gives 1-homogeneity.

Exercise 6.3

Show that the $\mathbb{K}$-vector space $\mathbb{K}^n$ of column vectors with $n$ entries is isomorphic to the $\mathbb{K}$-vector space $\mathbb{K}_n$ of row vectors with $n$ entries.

Solution

Taking the transpose induces a linear map $\mathbb{K}^n\to \mathbb{K}_n.$ Since for any $\vec v\in\mathbb{K}^n$ we have $(\vec v^T)^T=\vec v,$ this induced map is invertible and hence an isomorphism.

Exercise 6.4

Let $f : U \to V$ and $g: V \to W$ be linear maps. Show that \[\operatorname{rank}(g \circ f) \leqslant\min(\operatorname{rank}(f), \operatorname{rank}(g)).\]

Solution

Clearly $\operatorname{image}(g \circ f)$ is contained in $\operatorname{image}(g),$ so $\operatorname{rank}(g \circ f) \leqslant\operatorname{rank}(g).$ On the other hand, $\ker(g \circ f)$ clearly contains $\ker(f),$ so $\operatorname{nullity}(g \circ f) \geqslant\operatorname{nullity}(f),$ so the Rank-Nullity Theorem gives $\operatorname{rank}(g \circ f) = \dim U - \operatorname{nullity}(g \circ f) \leqslant\dim U - \operatorname{nullity}(f) = \operatorname{rank}(f).$

Exercise 6.5

Show that the $\mathbb{R}$-vector spaces $\mathsf{P}_n(\mathbb{R})$ and $\mathbb{R}^{n+1}$ are isomorphic for all $n \in \mathbb{N}.$

Solution

Let $n\in\mathbb{N}$ be arbitrary. Define the linear map $f:\mathbb{R}^{n+1}\to\mathsf{P}_n(\mathbb{R})$ by \[\begin{pmatrix} a_0\\ \vdots \\ a_{n} \end{pmatrix}\mapsto p:x\mapsto \sum_{k=0}^n a_kx^k.\] Injectivity of $f$: If two polynomials \[p:x\mapsto \sum_{k=0}^n a_kx^k\text{ and }q:x\mapsto \sum_{k=0}^nb_kx^k\] agree, then $p-q=0_{\mathsf{P}_n(\mathbb{R})}$ and hence \[\sum_{k=0}^n a_kx^k-\sum_{k=0}^n b_kx^k=\sum_{k=0}^n (a_k-b_k)x^k\] is the zero polynomial, which implies $a_k=b_k$ for all $k=0,\ldots,n$ and hence $f$ is injective.
Surjectivity of $f$: Let $p:x\mapsto \sum_{k=0}^na_kx^k,$ then \[f\left(\begin{pmatrix} a_0\\ \vdots \\ a_n\end{pmatrix}\right) = p.\]

Exercise 6.6

(requires concepts from M03 Analysis I) Let $C^\infty(\mathbb{R})$ be the $\mathbb{R}$-vector space of functions $\mathbb{R}\to \mathbb{R}$ which are infinitely often differentiable. Show that the map $D$ defined by \[D(f)(x) = f''(x) - 2 f'(x) + f(x)\] is a linear map $C^\infty(\mathbb{R}) \to C^\infty(\mathbb{R}).$ Is $D$ surjective? What is the dimension of $\operatorname{Ker}D$?

[Hint: show that $D(f)(x) = e^x \tfrac{d^2}{dx^2}(e^{-x} f(x)).$]

Solution

The identity $D(f)(x) = e^x \tfrac{d^2}{dx^2}(e^{-x} f(x))$ is easily verified using the product rule for differentiation (which is valid since $e^{-x}$ and $f(x)$ are both differentiable functions).

Since multiplying by $e^x$ is a bijective linear map from $C^\infty(\mathbb{R})$ to itself, it suffices to the kernel and image of the map $\tfrac{d^2}{dx^2}.$

Now we use the Fundamental Theorem of Calculus (Theorem 7.10 in the Analysis I script) to show that every continuous function $f$ is the derivative of a differentiable function $F,$ and this anti-derivative $F$ is unique up to a constant. Clearly, if $f$ is infinitely differentiable, so must be $F.$ Thus $\tfrac{d}{dx}$ is surjective as a linear map $C^\infty(\mathbb{R}) \to C^\infty(\mathbb{R}),$ with 1-dimensional kernel spanned by the constant 1. Repeating the process, $\tfrac{d^2}{dx^2}$ is also surjective, with 2-dimensional kernel spanned by 1 and any antiderivative of 1; clearly we can choose this to be $x \mapsto x.$

Thus our original $D$ is surjective and its kernel is 2-dimensional spanned by the functions $f_1(x) = e^x$ and $f_2(x) = x e^x.$

6 Linear maps

6.1 Linear maps

Examples

First properties of linear maps

Isomorphisms

6.2 Images, preimages, kernels

Linear maps and dimension

6.3 The rank-nullity theorem

Exercises

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