Linear Algebra I

3.3 Vector subspaces and isomorphisms

3.3.1 Vector subspaces

A vector subspace of a vector space is a subset that is itself a vector space, more precisely:

Definition 3.21 • Vector subspace

Let \(V\) be a \(\mathbb{K}\)-vector space. A subset \(U\subset V\) is called a vector subspace of \(V\) if \(U\) is non-empty and if \[\tag{3.8} s_1\cdot_Vv_1+_Vs_2\cdot_V v_2 \in U\quad \text{for all}\; s_1,s_2 \in \mathbb{K}\; \text{and all}\; v_1,v_2 \in U.\]

Remark 3.22

  1. Observe that since \(U\) is non-empty, it contains an element, say \(u.\) Since \(0\cdot_V u =0_V \in U\) it follows that the zero vector \(0_V\) lies in \(U.\) A vector subspace \(U\) is itself a vector space when we take \(0_U=0_V\) and borrow vector addition and scalar multiplication from \(V.\) Indeed, all of the properties in Definition 3.1 of \(+_V\) and \(\cdot_V\) hold for all elements of \(V\) and all scalars, hence also for all elements of \(U\subset V\) and all scalars. We only need to verify that we cannot fall out of \(U\) by vector addition and scalar multiplication, but this is precisely what the condition (3.8) states.

  2. A vector subspace is also called a linear subspace or simply a subspace.

The prototypical example of a vector subspace are lines and planes through the origin in \(\mathbb{R}^3\):

Example 3.23 • Lines through the origin

Let \(\vec{w}\neq 0_{\mathbb{R}^3},\) then the line \[U=\{s\vec{w}\, |\, s\in \mathbb{R}\} \subset \mathbb{R}^3\] is a vector subspace. Indeed, taking \(s=0\) it follows that \(0_{\mathbb{R}^3} \in U\) so that \(U\) is non-empty. Let \(\vec{u}_1,\vec{u}_2\) be vectors in \(U\) so that \(\vec{u}_1=t_1\vec{w}\) and \(\vec{u}_2=t_2\vec{w}\) for scalars \(t_1,t_2 \in \mathbb{R}.\) Let \(s_1,s_2 \in \mathbb{R},\) then \[s_1\vec{u}_1+s_2\vec{u}_2=s_1t_1\vec{w}+s_2t_2\vec{w}=\left(s_1t_1+s_2t_2\right)\vec{w} \in U\] so that \(U\subset \mathbb{R}^3\) is a subspace.

Example 3.24 • Zero vector space Let \(V\) be a \(\mathbb{K}\)-vector space and \(U=\{0_V\}\) the zero vector space arising from \(0_V.\) Then, by Definition 3.21 and the properties of Definition 3.1, it follows that \(U\) is a vector subspace of \(V.\)
Example 3.25 • Periodic functions Taking \(I=\mathbb{R}\) and \(\mathbb{K}=\mathbb{R}\) in Example 3.5, we see that the functions \(f : \mathbb{R}\to \mathbb{R}\) form an \(\mathbb{R}\)-vector space \(V=\mathsf{F}(\mathbb{R},\mathbb{R}).\) Consider the subset \[U=\left\{f \in \mathsf{F}(\mathbb{R},\mathbb{R})\,|\, f\;\text{is periodic with period}\; 2\pi\right\}\] consisting of \(2\pi\)-periodic functions, that is, an element \(f \in U\) satisfies \(f(x+2\pi)=f(x)\) for all \(x \in \mathbb{R}.\) Notice that \(U\) is not empty, as \(\cos : \mathbb{R}\to \mathbb{R}\) and \(\sin : \mathbb{R}\to \mathbb{R}\) are elements of \(U.\) Suppose \(f_1,f_2 \in U\) and \(s_1,s_2 \in \mathbb{R}.\) Then, we have for all \(x \in \mathbb{R}\) \[\begin{aligned} (s_1f_1+s_2f_2)(x+2\pi)&=s_1f_1(x+2\pi)+s_2f_2(x+2\pi)=s_1f_1(x)+s_2f_2(x)\\ &=(s_1f_1+s_2f_2)(x) \end{aligned}\] showing that \(s_1f_1+s_2f_2\) is periodic with period \(2\pi.\) By Definition 3.21, it follows that \(U\) is a vector subspace of \(\mathsf{F}(\mathbb{R},\mathbb{R}).\)

Recall, if \(\mathcal{X},\mathcal{W}\) are sets, \(\mathcal{Y}\subset \mathcal{X},\) \(\mathcal{Z}\subset \mathcal{W}\) subsets and \(f : \mathcal{X} \to \mathcal{W}\) a mapping, then the image of \(\mathcal{Y}\) under \(f\) is the set \[f(\mathcal{Y})=\left\{w \in \mathcal{W}\,|\, \text{there exists an element}\; y \in \mathcal{Y}\;\text{with}\; f(y)=w\right\}\] consisting of all the elements in \(\mathcal{W}\) which are hit by an element of \(\mathcal{Y}\) under the mapping \(f.\) In the special case where \(\mathcal{Y}\) is all of \(\mathcal{X},\) that is, \(\mathcal{Y}=\mathcal{X},\) it is also customary to write \(\operatorname{Im}(f)\) instead of \(f(\mathcal{X})\) and simply speak of the image of \(f.\) Similarly, the preimage of \(\mathcal{Z}\) under \(f\) is the set \[f^{-1}(\mathcal{Z})=\left\{x \in \mathcal{X}\,|\,f(x) \in \mathcal{Z}\right\}\] consisting of all the elements in \(\mathcal{X}\) which are mapped onto elements of \(\mathcal{Z}\) under \(f.\) Notice that \(f\) is not assumed to be bijective, hence the inverse mapping \(f^{-1} : \mathcal{W} \to \mathcal{X}\) does not need to exist (and in fact the definition of the preimage does not involve the inverse mapping). Nonetheless the notation \(f^{-1}(\mathcal{Z})\) is customary.

It is natural to ask how the image and preimage of subspaces look like under a linear map:

Proposition 3.26

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 3.15, \(f(0_V)=0_W,\) 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 3.21 again, we conclude that \(f(U)\) is a subspace of \(W.\) The second claim is left to the reader as an exercise.

Vector subspaces are stable under intersection in the following sense:

Proposition 3.27

Let \(V\) be a \(\mathbb{K}\)-vector space, \(n\geqslant 2\) a natural number and \(U_1,\ldots,U_n\) vector subspaces of \(V.\) Then the intersection \[U^{\prime}=\bigcap_{j=1}^n U_j=\left\{v \in V\,|\, v \in U_j\;\text{for all}\; j=1,\ldots,n\right\}\] is a vector subspace of \(V\) as well.

Proof. Since \(U_j\) is a vector subspace, \(0_V \in U_j\) for all \(j=1,\ldots,n.\) Therefore, \(0_V \in U^{\prime},\) hence \(U^{\prime}\) is not empty. Let \(u_1,u_2 \in U^{\prime}\) and \(s_1,s_2 \in \mathbb{K}.\) By assumption, \(u_1,u_2 \in U_j\) for all \(j=1,\ldots,n.\) Since \(U_j\) is a vector subspace for all \(j=1,\ldots,n\) it follows that \(s_1u_1+s_2 u_2 \in U_j\) for all \(j=1,\ldots,n\) and hence \(s_1u_1+s_2u_2 \in U^{\prime}.\) By Definition 3.21, it follows that \(U^{\prime}\) is a vector subspace of \(V.\)
Remark 3.28

Notice that the union of subspaces need not be a subspace. Let \(V=\mathbb{R}^2,\) \(\{\vec{e}_1,\vec{e}_2\}\) its standard basis and \[U_1=\left\{s\vec{e}_1\,|\,s\in \mathbb{R}\right\} \quad \text{and} \quad U_2=\left\{s\vec{e}_2\,|\,s\in \mathbb{R}\right\}.\] Then \(\vec{e}_1 \in U_1\cup U_2\) and \(\vec{e}_2\in U_1\cup U_2,\) but \(\vec{e}_1+\vec{e}_2 \notin U_1\cup U_2.\)

The kernel of a linear map \(f : V \to W\) consists of those vectors in \(V\) that are mapped onto the zero vector of \(W\):

Definition 3.29 • 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 3.30

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.

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

Lemma 3.31

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 3.15, 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 \(v_1=v_2.\)
An immediate consequence of Proposition 3.26 is:
Corollary 3.32

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 \(V.\)

3.3.2 Isomorphisms

Definition 3.33 • 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.

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 3.31 gives:
Proposition 3.34

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

3.4 Generating sets

Definition 3.35 • Linear combination

Let \(V\) be a \(\mathbb{K}\)-vector space, \(k \in \mathbb{N}\) and \(\{v_1,\ldots, v_k\}\) a set of vectors from \(V.\) A linear combination of the vectors \(\{v_1,\ldots,v_k\}\) is a vector of the form \[w=s_1v_1+\cdots+s_kv_k=\sum_{i=1}^ks_i v_i\] for some \(s_1,\ldots,s_k \in \mathbb{K}.\)

Example 3.36

For \(n \in \mathbb{N}\) with \(n\geqslant 2\) consider \(V=\mathsf{P}_n(\mathbb{R})\) and the polynomials \(p_1,p_2,p_3 \in \mathsf{P}_n(\mathbb{R})\) defined by the rules \(p_1(x)=1,\) \(p_2(x)=x,\) \(p_3(x)=x^2\) for all \(x \in \mathbb{R}.\) A linear combination of \(\{p_1,p_2,p_3\}\) is a polynomial of the form \(p(x)=ax^2+bx+c\) where \(a,b,c \in \mathbb{R}.\)

Definition 3.37 • Subspace generated by a set

Let \(V\) be a \(\mathbb{K}\)-vector space and \(\mathcal{S}\subset V\) be a non-empty subset. The subspace generated by \(\mathcal{S}\) is the set \(\operatorname{span}(\mathcal{S})\) whose elements are linear combinations of finitely many vectors in \(\mathcal{S}.\) The set \(\operatorname{span}(\mathcal{S})\) is called the span of \(\mathcal{S}\). Formally, we have \[\operatorname{span}(\mathcal{S})=\left\{v \in V\,\Big|\, v=\sum_{i=1}^ks_iv_i,k \in \mathbb{N}, s_1,\ldots,s_k \in \mathbb{K},v_1,\ldots,v_k \in \mathcal{S}\right\}.\]

Remark 3.38

The notation \(\langle \mathcal{S}\rangle\) for the span of \(\mathcal{S}\) is also in use.

Proposition 3.39

Let \(V\) be a \(\mathbb{K}\)-vector space and \(\mathcal{S}\subset V\) be a non-empty subset. Then \(\operatorname{span}(\mathcal{S})\) is a vector subspace of \(V.\)

Proof. Since \(\mathcal{S}\) is non-empty it contains some element, say \(u.\) Since \(u\) itself is a linear combination of \(\{u\},\) it follows that \(\operatorname{span}(\mathcal{S})\) is non-empty. Let \(k\in \mathbb{N}\) and \(v_1=t_1w_1+\cdots+t_kw_k\) for \(t_1,\ldots t_k \in \mathbb{K}\) and \(w_1,\ldots, w_k \in \mathcal{S}\) be a linear combination of vectors in \(\mathcal{S}.\) Furthermore, let \(j \in \mathbb{N}\) and \(v_2=\hat{t}_1\hat{w}_1+\cdots+\hat{t}_j\hat{w}_j\) for \(\hat{t}_1,\ldots,\hat{t}_j\) and \(\hat{w}_1,\ldots,\hat{w}_j \in \mathcal{S}\) be another linear combination of vectors in \(\mathcal{S}.\) By Definition 3.21, it suffices to show that for all \(s_1,s_2 \in \mathbb{K}\) the vector \(s_1v_1+s_2 v_2\) is a linear combination of vectors in \(\mathcal{S}.\) Since \[\begin{aligned} s_1v_1+s_2v_2&=s_1(t_1w_1+\cdots+t_kw_k)+s_2(\hat{t}_1\hat{w}_1+\cdots+\hat{t}_j\hat{w}_j)\\ &=s_1t_1w_1+\cdots+s_1t_kw_k+s_2\hat{t}_1\hat{w}_1+\cdots+s_2\hat{t}_j\hat{w}_j \end{aligned}\] is a linear combination of the vectors \(\{w_1,\ldots,w_k,\hat{w}_1,\ldots,\hat{w}_j\}\) in \(\mathcal{S},\) the claim follows.
Remark 3.40

For a subset \(\mathcal{S}\subset V,\) we may alternatively define \(\operatorname{span}(\mathcal{S})\) to be the smallest vector subspace of \(V\) that contains \(\mathcal{S}.\) This has the advantage of \(\mathcal{S}\) being allowed to be empty, in which case \(\operatorname{span}(\emptyset)=\{0_V\},\) that is, the empty set is a generating set for the zero vector space.

Definition 3.41

Let \(V\) be a \(\mathbb{K}\)-vector space. A subset \(\mathcal{S}\subset V\) is called a generating set if \(\operatorname{span}(\mathcal{S}) =V.\) The vector space \(V\) is called finite dimensional if \(V\) admits a generating set with finitely many elements (also called a finite set). A vector space that is not finite dimensional will be call infinite dimensional.

Example 3.42

Thinking of a field \(\mathbb{K}\) as a \(\mathbb{K}\)-vector space, the set \(\mathcal{S}=\{1_{\mathbb{K}}\}\) consisting of the identity element of multiplication is a generating set for \(V=\mathbb{K}.\) Indeed, for every \(x \in \mathbb{K}\) we have \(x=x\cdot_{V}1_{\mathbb{K}}.\)

Example 3.43

The standard basis \(\mathcal{S}=\{\vec{e}_1,\ldots,\vec{e}_n\}\) is a generating set for \(\mathbb{K}^n,\) since for all \(\vec{x}=(x_i)_{1\leqslant i\leqslant n} \in \mathbb{K}^n,\) we can write \(\vec{x}=x_1\vec{e}_1+\cdots+x_n\vec{e}_n\) so that \(\vec{x}\) is a linear combination of elements of \(\mathcal{S}.\)

Example 3.44

Let \(\mathbf{E}_{k,l} \in M_{m,n}(\mathbb{K})\) for \(1\leqslant k \leqslant m\) and \(1\leqslant l \leqslant n\) denote the \(m\)-by-\(n\) matrix satisfying \(\mathbf{E}_{k,l}=(\delta_{ki}\delta_{lj})_{1\leqslant i\leqslant m, 1\leqslant j \leqslant n}.\) For example, for \(m=2\) and \(n=3\) we have \[\mathbf{E}_{1,1}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},\qquad \mathbf{E}_{1,2}=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix},\qquad \mathbf{E}_{1,3}=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\] and \[\mathbf{E}_{2,1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},\qquad \mathbf{E}_{2,2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\qquad \mathbf{E}_{2,3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}.\] Then \(\mathcal{S}=\{\mathbf{E}_{k,l}\}_{1\leqslant k\leqslant m, 1\leqslant l \leqslant n}\) is a generating set for \(M_{m,n}(\mathbb{K}),\) since a matrix \(\mathbf{A}\in M_{m,n}(\mathbb{K})\) can be written as \[\mathbf{A}=\sum_{k=1}^m\sum_{l=1}^n A_{kl}\mathbf{E}_{k,l}\] so that \(\mathbf{A}\) is a linear combination of the elements of \(\mathcal{S}.\)

Example 3.45

The vector space \(\mathsf{P}(\mathbb{R})\) of polynomials is infinite dimensional. In order to see this, consider a finite set of polynomials \(\{p_1,\ldots,p_n\},\) \(n \in \mathbb{N}\) and let \(d_i\) denote the degree of the polynomial \(p_i\) for \(i=1,\ldots,n.\) We set \(D =\max\{d_1,\ldots,d_n\}.\) Since a linear combination of the polynomials \(\{p_1,\ldots,p_n\}\) has degree at most \(D ,\) any polynomial \(q\) whose degree is strictly larger than \(D\) will satisfy \(q \notin \operatorname{span}\{p_1,\ldots,p_n\}.\) It follows that \(\mathsf{P}(\mathbb{R})\) cannot be generated by a finite set of polynomials.

Lemma 3.46

Let \(f : V\to W\) be linear and \(\mathcal{S}\subset V\) a generating set. If \(f\) is surjective, then \(f(\mathcal{S})\) is a generating set for \(W.\) Furthermore, if \(f\) is bijective, then \(V\) is finite dimensional if and only if \(W\) is finite dimensional.

Proof. 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 \(W=\operatorname{span}(f(\mathcal{S})).\)

For the second claim suppose \(V\) is finite dimensional, hence we have a finite set \(\mathcal{S}\) with \(\operatorname{span}(\mathcal{S})=V.\) The set \(f(\mathcal{S})\) is finite as well and satisfies \(\operatorname{span}( f(\mathcal{S}))=W\) by the previous argument, hence \(W\) is finite dimensional as well. Conversely suppose \(W\) is finite dimensional with generating set \(\mathcal{T}\subset W.\) Since \(f\) is bijective there exists an inverse mapping \(f^{-1} : W \to V\) which is surjective, hence \(V=\operatorname{span}(f^{-1}(\mathcal{T}))\) so that \(V\) is finite dimensional as well.

3.5 Linear independence and bases

A set of vectors where no vector can be expressed as a linear combination of the other vectors is called linearly independent. More precisely:

Definition 3.47 • Linear independence

Let \(\mathcal{S}\subset V\) be a non-empty finite subset so that \(\mathcal{S}=\{v_1,\ldots ,v_k\}\) for distinct vectors \(v_i \in V,\) \(i=1,\ldots,k.\) We say \(\mathcal{S}\) is linearly independent if \[s_1v_1+\cdots+s_kv_k=0_V \qquad \iff \qquad s_1=\cdots=s_k=0,\] where \(s_1,\ldots,s_k \in \mathbb{K}.\) If \(\mathcal{S}\) is not linearly independent, then \(\mathcal{S}\) is called linearly dependent. Furthermore, we call a subset \(\mathcal{S}\subset V\) linearly independent if every finite subset of \(\mathcal{S}\) is linearly independent. We will call distinct vectors \(v_1,\ldots,v_k\) linearly independent/dependent if the set \(\{v_1,\ldots,v_k\}\) is linearly independent/dependent.

Remark 3.48

Instead of distinct, many authors write pairwise distinct, which means that all pairs of vectors \(v_i,v_j\) with \(i\neq j\) satisfy \(v_i\neq v_j.\) Of course, this simply means that the list \(v_1,\ldots,v_k\) of vectors is not allowed to contain a vector more than once.

Notice that if the vectors \(v_1,\ldots,v_k \in V\) are linearly dependent, then there exist scalars \(s_1,\ldots,s_k,\) not all zero, so that \(\sum_{i=1}^ks_iv_i=0_V.\) After possibly changing the numbering of the vectors and scalars, we can assume that \(s_1\neq 0.\) Therefore, we can write \[v_1=-\sum_{i=2}^k\left(\frac{s_i}{s_1}\right)v_i,\] so that \(v_1\) is a linear combination of the vectors \(v_2,\ldots,v_k.\)

Also, observe that a subset \(\mathcal{T}\) of a linearly independent set \(\mathcal{S}\) is itself linearly independent. (Why?)

Example 3.49

We consider the polynomials \(p_1,p_2,p_3 \in \mathsf{P}(\mathbb{R})\) defined by the rules \(p_1(x)=1,p_2(x)=x,p_3(x)=x^2\) for all \(x \in \mathbb{R}.\) Then \(\{p_1,p_2,p_3\}\) is linearly independent. In order to see this, consider the condition \[\tag{3.9} s_1p_1+s_2p_2+s_3p_3=0_{\mathsf{P}(\mathbb{R})}=o\] where \(o : \mathbb{R}\to \mathbb{R}\) denotes the zero polynomial. Since (3.9) means that \[s_1p_1(x)+s_2p_2(x)+s_3p_3(x)=o(x),\] for all \(x \in \mathbb{R},\) we can evaluate this condition for any choice of real number \(x.\) Taking \(x=0\) gives \[s_1p_1(0)+s_2p_2(0)+s_3p_3(0)=o(0)=0=s_1.\] Taking \(x=1\) and \(x=-1\) gives \[\begin{aligned} 0&=s_2p_2(1)+s_3p_3(1)=s_2+s_3,\\ 0&=s_2p_2(-1)+s_3p_3(-1)=-s_2+s_3, \end{aligned}\] so that \(s_2=s_3=0\) as well. It follows that \(\{p_1,p_2,p_3\}\) is linearly independent.

Remark 3.50

By convention, the empty set is linearly independent.

Definition 3.51 • Basis

A subset \(\mathcal{S}\subset V\) which is a generating set of \(V\) and also linearly independent is called a basis of \(V.\)

Example 3.52 Thinking of a field \(\mathbb{K}\) as a \(\mathbb{K}\)-vector space, the set \(\{1_{\mathbb{K}}\}\) is linearly independent, since \(1_{\mathbb{K}}\neq 0_{\mathbb{K}}.\) Example 3.42 implies that \(\{1_{\mathbb{K}}\}\) is a basis of \(\mathbb{K}.\)
Example 3.53 Clearly, the standard basis \(\{\vec{e}_1,\ldots,\vec{e}_n\}\) of \(\mathbb{K}^n\) is linearly independent since \[s_1\vec{e}_1+\cdots+s_n\vec{e}_n=\begin{pmatrix} s_1 \\ \vdots \\ s_n\end{pmatrix}=0_{\mathbb{K}^{n}}=\begin{pmatrix} 0 \\ \vdots \\ 0 \end{pmatrix} \qquad \iff \qquad s_1=\dots=s_n=0.\] It follows together with Example 3.43 that the standard basis of \(\mathbb{K}^n\) is indeed a basis in the sense of Definition 3.51.
Example 3.54 The matrices \(\mathbf{E}_{k,l} \in M_{m,n}(\mathbb{K})\) for \(1\leqslant k \leqslant m\) and \(1\leqslant l \leqslant n\) are linearly independent. Suppose we have scalars \(s_{kl} \in \mathbb{K}\) such that \[\sum_{k=1}^m\sum_{l=1}^ns_{kl}\mathbf{E}_{k,l}=\mathbf{0}_{m,n}=\begin{pmatrix}s_{11} & \cdots & s_{1n} \\ \vdots & \ddots & \vdots\\ s_{m1} & \cdots & s_{mn}\end{pmatrix}=\begin{pmatrix}0 & \cdots & 0 \\ \vdots & \ddots & \vdots\\ 0 & \cdots & 0\end{pmatrix}\] so \(s_{kl}=0\) for all \(1\leqslant k \leqslant m\) and all \(1\leqslant l \leqslant n.\) It follows together with Example 3.44 that \(\{\mathbf{E}_{k,l}\}_{1\leqslant k\leqslant m, 1\leqslant l \leqslant n}\) is a basis of \(M_{m,n}(\mathbb{K}).\) We refer to \(\{\mathbf{E}_{k,l}\}_{1\leqslant k\leqslant m, 1\leqslant l \leqslant n}\) as the standard basis of \(M_{m,n}(\mathbb{K})\).
Example 3.55 Combining Remark 3.40 and Remark 3.50 we conclude that the empty set is a basis for the zero vector space \(\{0\}.\)
Lemma 3.56

Let \(f : V \to W\) be an injective linear map. Suppose \(\mathcal{S}\subset V\) is linearly independent, then \(f(\mathcal{S}) \subset W\) is also linearly independent.

Proof. Let \(\{w_1,\ldots,w_k\}\subset f(\mathcal{S})\) be a finite subset for some \(k \in \mathbb{N}\) some and distinct vectors \(w_i \in W,\) where \(1\leqslant i\leqslant k.\) Then there exist vectors \(v_1,\ldots,v_k\) with \(f(v_i)=w_i\) for \(1\leqslant i\leqslant k.\) Suppose there exist scalars \(s_1,\ldots,s_k\) such that \(s_1w_1+\cdots+s_kw_k=0_W.\) Using the linearity of \(f,\) this implies \[0_W=s_1w_1+\cdots+s_kw_k=s_1f(v_1)+\cdots +s_kf(v_k)=f(s_1v_1+\cdots+s_kv_k).\] Since \(f\) is injective we have \(\operatorname{Ker}(f)=\{0_V\}\) by Lemma 3.31. Since \(s_1v_1+\cdots+s_k v_k \in \operatorname{Ker}{f}\) it follows that \(s_1v_1+\cdots+s_kv_k=0_V,\) hence \(s_1=\cdots=s_k=0\) by the linear independence of \(\mathcal{S}.\) It follows that \(f(\mathcal{S})\) is linearly independent as well.

Exercises

Exercise 3.57

Let \(U\subset V\) be a vector subspace and \(k \in \mathbb{N}\) with \(k\geqslant 2.\) Show that for \(u_1,\ldots,u_k \in U\) and \(s_1,\ldots,s_k \in \mathbb{K},\) we have \(s_1u_1+\cdots+s_ku_k \in U.\)

Solution

We will prove the claim by induction on \(k.\) If \(u_1,u_2\in U\) and \(s_1,s_2\in\mathbb{K},\) then \(s_1u_1+s_2u_2\in U\) according to Definition 3.21 and the statement is anchored. Inductive step: Assume \(k\geqslant 2\) and let \(u_1,\ldots,u_{k+1}\in U\) and \(s_1,\ldots,s_{k+1}\in\mathbb{K}.\) According to the induction hypothesis, \[u=\sum_{j=1}^k s_ju_j\in U\] and hence \[\sum_{j=1}^{k+1}s_ju_j = u + s_{k+1}u_{k+1}\in U\] again by Definition 3.21.
Exercise 3.58 • Planes through the origin

Let \(\vec{w}_1,\vec{w}_2\neq 0_{\mathbb{R}^3}\) and \(\vec{w}_1\neq s\vec{w}_2\) for all \(s\in \mathbb{R}.\) Show that the plane \[U=\{s_1\vec{w}_1+s_2\vec{w}_2\,|\, s_1,s_2 \in \mathbb{R}\}\] is a vector subspace of \(\mathbb{R}^3.\)

Solution

According to Definition 3.21 we first need to show that \(U\ne\emptyset\): By choosing \(s_1=s_2=0,\) we see that \(0_{\mathbb{R}^3}\in U\) and hence \(U\ne\emptyset.\) Let \(\vec u,\vec v\in U.\) We need to show that \(t_1\vec u+t_2\vec v\in U,\) for all \(t_1,t_2\in\mathbb{R}.\) There exist scalars \(s_1,\hat s_1, s_2,\hat s_2\) such that \(\vec u = s_1 \vec w_1 + s_2\vec w_2\) and \(\vec v = \hat s_1 \vec w_1+\hat s_2 \vec w_2\) and hence \[\begin{aligned}t_1\vec u + t_2\vec v & = t_1(s_1 \vec w_1 + s_2\vec w_2)+t_2(\hat s_1 \vec w_1+\hat s_2 \vec w_2)\\ & = (t_1s_1+t_2\hat s_1)\vec w_1 + (t_1s_2+t_2\hat s_2)\vec w_2\in U. \end{aligned}\] Note that we do not need the assumption \(\vec w_1\ne s\vec w_2\) for all \(s\in\mathbb{R}\) throughout the argument. Indeed, if \(\vec w_1 = s\vec w_2\) for some \(s\ne 0,\) then \(U\) is still a subspace of \(\mathbb{R}^3\) but in this case, it would be called a line instead of a plane.
Exercise 3.59 • Polynomials

Let \(n \in \mathbb{N}\cup\{0\}\) and \(\mathsf{P}_n(\mathbb{R})\) denote the subset of \(\mathsf{P}(\mathbb{R})\) consisting of polynomials of degree at most \(n\). Show that \(\mathsf{P}_n(\mathbb{R})\) is a subspace of \(\mathsf{P}(\mathbb{R})\) for all \(n \in \mathbb{N}\cup\{0\}.\)

Solution

Let \(n\in \mathbb{N}\cup\{0\}\) be arbitrary. The constant polynomial \(p:\mathbb{R}\to\mathbb{R}, x\mapsto 0_\mathbb{R}\) is an element of \(\mathsf{P}_n(\mathbb{R})\) and hence, \(\mathsf{P}_n(\mathbb{R})\ne\emptyset.\) Let now \(p,q\in \mathsf{P}_n(\mathbb{R})\) be defined by \[\begin{aligned} p&: x\mapsto \sum_{k=0}^n a_kx^k\\ q&: x\mapsto \sum_{k=0}^n b_kx^k, \end{aligned}\] where \(a_0,b_0,a_1,b_1,\ldots,a_n,b_n\in\mathbb{R}.\) For \(s_1,s_2\in \mathbb{R},\) the polynomial \(s_1p+s_2q\) is given by the function \[x\mapsto \sum_{k=0}^n (s_1a_k+s_2b_k)x^k\] and hence is an element of \(\mathsf{P}_n(\mathbb{R})\) and hence \(\mathsf{P}_n(\mathbb{R})\) is a subspace of \(\mathsf{P}(\mathbb{R})\) according to Definition 3.21.
Exercise 3.60

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 3.61

Show that the \(\mathbb{R}\)-vector spaces \(\mathsf{P}_n(\mathbb{R})\) and \(\mathbb{R}^{n+1}\) are isomorphic for all \(n \in \mathbb{N}\cup\{0\}.\)

Solution

Let \(n\in\mathbb{N}\cup\{0\}\) 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 3.62 Show that for a non-empty subset \(\mathcal{S}\) of a \(\mathbb{K}\)-vector space \(V,\) the set \(\operatorname{span}(\mathcal{S})\) as defined in Definition 3.37 is the same as the set \(\operatorname{span}(\mathcal{S})\) as defined in Remark 3.40. In particular, Proposition 3.39 remains true when removing the assumption that \(\mathcal{S}\) is non-empty.

Solution

Let \(\mathcal S\subset V\) be a non-empty set. We need to show that \(\operatorname{span}(\mathcal S)\) as defined in Definition 3.37 is the smallest subspace of \(V\) that contains \(\mathcal S.\) To this end, let \(U\) be any subspace which contains \(\mathcal S.\) We will show that \(\operatorname{span}(\mathcal S)\subset U.\) Since \(\mathcal S\subset U\) and \(U\) is a subspace, \(U\) contains every finite linear combination of elements in \(\mathcal S\) (see Exercise 3.57). The set \(\operatorname{span}(\mathcal S)\) is by Definition 3.37 the set of all finite linear combinations of elements of \(\mathcal S\) and therefore \(U\supset \operatorname{span}(\mathcal S)\) and hence \(\operatorname{span}(\mathcal S)\) is the smallest subspace of \(V\) that contains \(\mathcal S.\)
Exercise 3.63

Show that a subset \(\{v\}\) consisting of a single vector \(v \in V\) is linearly independent if and only if \(v\neq 0_V.\)

Solution

The subset \(\{v\}\subset V\) is linearly independent if \(sv = 0_V\Longleftrightarrow s= 0.\) If \(v=0_V,\) then \(sv=0_V\) for any \(s\in\mathbb{K}\) and therefore, the set is linearly dependent. If \(v\ne 0_V,\) then \(sv=0_V\) which implies \(s=0\) by the last item of Lemma 3.9 and hence \(\{v\}\) is linearly independent.
Linear Algebra I

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