Exercise. Determine all vector spaces that have exactly one basis, up to reordering, and prove that no others exist. Note that this is different from the original question.
Theorem. A vector space \(V\) over \(\mathbb{F}\) has exactly one basis, up to reordering, if and only if it satisfies one of the two following statements:
1. \(V=\left\{ 0\right\}\) over any field \(\mathbb{F}\)
2. \(\mathbb{F}\) is isomorphic to \(\mathbb{Z}_{2}\) and \(\exists0\ne v\text{ s.t. }V=\left\{ 0,v\right\}\).
Remark. In the book, a basis is defined (\(1.8\)) as a list, which is a finite, ordered object.
Therefore, exactly one basis, up to reordering, means that every basis of the vector space is a permutation of a single list.
Remark. If you did not learn about isomorphism and prime fields, you can think of “\(\mathbb{F}\) is isomorphic to \(\mathbb{Z}_{2}\)” as \(\mathbb{F}=\{0,1\}\) with \(1+1=0\).
Lemma 1. If \(V\) isn’t finite-dimensional, then V has no basis. In particular, V does not have exactly one basis, up to reordering.
Proof. According to definition \(2.9\), a vector space is finite-dimensional iff some list of vectors in it spans the space.
Therefore, if it isn’t finite-dimensional (and by definition \(2.13\) it is then called infinite-dimensional), no list of vectors in it spans the space.
Since by definition \(2.26\) a basis must span the space, no list of vectors can be a basis of the space, hence it has no bases and in particular doesn’t have exactly one. ∎
Lemma 2. \(V=\left\{ 0\right\}\) over any field \(\mathbb{F}\) has exactly one basis.
Proof. \(\left\{ 0\right\}\) is a vector space over any field \(\mathbb{F}\) - after definition \(1.22\), \(\left\{ 0\right\}\) is mentioned as the simplest vector space.
Definition \(2.4\) asserts \(\text{Span }()=\left\{ 0\right\}\).
Definition \(2.15\) asserts that the empty list is linearly independent.
Therefore, by definition \(2.26\), \(\left(\right)\) is a basis for \(\left\{ 0\right\}\).
Any non-empty list of vectors in \(\left\{ 0\right\}\) must contain the zero vector, which makes it linearly dependent (according to example \(2.18\)).
Thus, no non-empty list is a basis of \(\left\{ 0\right\}\) (by definition \(2.26\)), so \(\left(\right)\) is its only basis. ∎
Lemma 3. Let \(V\) be a finite-dimensional vector space over field \(\mathbb{F}\), and let \(\mathcal{B}=\left(v_{1},\dots,v_{n}\right)\) be a basis of \(V\) such that \(n=1\).
Then \(V\) has exactly one basis, up to reordering, if and only if \(\mathbb{F}\) is isomorphic to \(\mathbb{Z}_{2}\).
Proof. If \(\mathbb{F}\) is isomorphic to \(\mathbb{Z}_{2}\), then \(\mathbb{F}=\left\{ 0,1\right\}\) with \(1+1=0\), and \[V=\text{Span }\mathcal{B}=\left\{ av_{1}\mid a\in\mathbb{F}\right\} =\left\{ 0,v_{1}\right\}\] and there is no other basis other than \(\mathcal{B}=\left(v_{1}\right)\):
Since \(\mathcal{B}\) is linearly independent, \(v_{1}\ne0\) (by example \(2.18\)).
The empty list is not a basis, since by definition \(2.4\), \(\text{Span }\left(\right)=\left\{ 0\right\} \ne V\).
Every vector in \(V\) is \(0\) or \(v_{1}\), so every list of vectors in \(V\) would contain some combination of these two elements.
By example \(2.18\), any list containing \(0\), or containing \(v_{1}\) more than once, is linearly dependent and therefore not a basis.
Hence, overall, the only possible basis remains \(\left(v_{1}\right)\).
If \(\mathbb{F}\) is not isomorphic to \(\mathbb{Z}_{2}\), then \(\mathbb{F}\) has more than two elements, since every field with exactly two elements is isomorphic to \(\mathbb{Z}_{2}\). Hence, there exists \(c\in\mathbb{F}\) with \(c\notin\{0,1\}\).
Because \(c\notin\left\{ 0,1\right\}\) and \(v_{1}\) cannot be \(0\) (\(\mathcal{B}\) would be linearly dependent), \(cv_{1}\ne v_{1}\):
Otherwise, \[0=v_{1}-cv_{1}\iff0=\left(1-c\right)v_{1}\underset{\text{Exercises 1B Q 2}}{\iff}1-c=0\iff c=1\] which is a contradiction.
Hence, \(\left\{ cv_{1}\right\} \ne\left\{ v_{1}\right\}\), and we will show that \(\mathcal{B}'=\left(cv_{1}\right)\) is also a basis for \(V\) (we have shown that they are different up to reordering).
Because \(v_{1}\) isn’t the zero vector and \(c\ne0\), the contra-positive of exercises \(1B\) question \(2\) gives \(cv_{1}\ne0\), and \(\mathcal{B}'\) contains a single non-zero vector, hence it is linearly independent.
Additionally, \[V=\text{Span }\mathcal{B}=\left\{ av_{1}\mid a\in\mathbb{F}\right\} =\left\{ a\cdot1\cdot v_{1}\mid a\in\mathbb{F}\right\} \underset{0\ne c\in\mathbb{F}}{=}\left\{ a\cdot\frac{1}{c}\cdot c\cdot v_{1}\mid a\in\mathbb{F}\right\} =\left\{ \frac{a}{c}\cdot cv_{1}\mid a\in\mathbb{F}\right\} \subseteq\text{Span }\mathcal{B}'\] and since \(\text{Span }\mathcal{B}'\) is a subspace of \(V\) (by \(2.6\)), the two sided inclusion means \(\mathcal{B}'\) spans \(V\), and therefore it is a basis for \(V\).
This means that if \(\mathbb{F}\) is not isomorphic to \(\mathbb{Z}_{2}\), \(V\) has more than one basis, up to reordering. ∎
Lemma 4. Let \(V\) be a finite-dimensional vector space over field \(\mathbb{F}\), and let \(\mathcal{B}=\left(v_{1},\dots,v_{n}\right)\) be a basis of \(V\) such that \(n\ge2\). Then \(\mathcal{B}\) is not the only basis of \(V\), up to reordering.
Proof. Let us look at \(\mathcal{B}'=\left(v_{1}+v_{2},v_{2},\dots,v_{n}\right)\) and let \(v\in V\).
According to definition \(2.26\), since \(\mathcal{B}\) is a basis of \(V\), \(\mathcal{B}\) spans \(V\).
Therefore, \[\exists a_{1},a_{2},a_{3},\dots,a_{n}\in\mathbb{F}\ \text{ s.t. }v=a_{1}v_{1}+a_{2}v_{2}+a_{3}v_{3}+\dots+a_{n}v_{n}\]
We can select \[\begin{cases} c_{1}=a_{1} & \in\mathbb{F}\\ c_{2}=a_{2}-a_{1} & \in\mathbb{F}\\ c_{3}=a_{3} & \in\mathbb{F}\\ \vdots\\ c_{n}=a_{n} & \in\mathbb{F} \end{cases}\]
we get \[v=a_{1}v_{1}+a_{2}v_{2}+a_{3}v_{3}+\dots+a_{n}v_{n}=a_{1}v_{1}+a_{1}v_{2}+a_{2}v_{2}-a_{1}v_{2}+a_{3}v_{3}+\dots+a_{n}v_{n}\] \[=c_{1}\left(v_{1}+v_{2}\right)+c_{2}v_{2}+c_{3}v_{3}+\dots+c_{n}v_{n}\in\text{Span }\mathcal{B}'\] Hence \(\mathcal{B}'\) also spans \(V\).
Let \(b_{1},b_{2},b_{3},\dots,b_{n}\in\mathbb{F}\) s.t. \[b_{1}\left(v_{1}+v_{2}\right)+b_{2}v_{2}+b_{3}v_{3}+\dots+b_{n}v_{n}=0\]
Then \[b_{1}v_{1}+\left(b_{1}+b_{2}\right)v_{2}+\dots+b_{n}v_{n}=0\]
According to definition \(2.26\), since \(\mathcal{B}\) is a basis of \(V\), \(\mathcal{B}\) is linearly independent and \[b_{1}=0,\quad\left(b_{1}+b_{2}\right)=0,\quad\dots,\quad b_{n}=0\] hence \[b_{1}=b_{2}=\dots=b_{n}=0\]
and according to definition \(2.15\), \(\mathcal{B}'\) is linearly independent as well.
According to definition \(2.26\), \(\mathcal{B}'\) is a basis of \(V\).
\(\mathcal{B}'\ne\mathcal{B}\), nor is it a reordering of \(\mathcal{B}\): let us assume for the sake of contradiction that \(v_{1}+v_{2}\in\mathcal{B}\).
Then \(\exists k\in\left\{ 1,\dots,n\right\} \text{ s.t. }v_{k}=v_{1}+v_{2}\).
If \(k=1\) or \(k=2\) then accordingly \(v_{1}+v_{2}=v_{1}\) or \(v_{1}+v_{2}=v_{2}\) hence \(v_{2}\) or \(v_{1}\) is the zero vector, contradicting the linear independence of \(\mathcal{B}\) (according to example \(2.18\), any list containing the zero vector is linearly dependent).
If \(k\ge3\) then \(\mathcal{B}\) is of the form \(\left(v_{1},v_{2},\dots,v_{1}+v_{2},\dots,v_{n}\right)\) and that \[1\cdot v_{1}+1\cdot v_{2}+0\cdot v_{3}+\dots+0\cdot v_{k-1}+\left(-1\right)\cdot v_{k}+0\cdot v_{k+1}+\dots+0\cdot v_{n}=0\] contradicting the linear independence of \(\mathcal{B}\).
Hence \(\mathcal{B}'\) contains a vector not in \(\mathcal{B}\), so \(\mathcal{B}'\) is a basis of \(V\) different from \(\mathcal{B}\), up to reordering. ∎
Proof. Tying it all together,
Let \(V\) be a vector space over field \(\mathbb{F}\).
\(\Leftarrow\): Let us assume \(V\) satisfies one of the two following statements:
1. \(V=\left\{ 0\right\}\) over any field \(\mathbb{F}\)
2. \(\mathbb{F}\) is isomorphic to \(\mathbb{Z}_{2}\) and \(\exists0\ne v\text{ s.t. }V=\left\{ 0,v\right\}\).
In the first case, we proved in lemma 2 that it has exactly one basis, up to reordering.
In the second case, \(\left(v\right)\) is a basis of \(V\) and we proved in lemma 3 that \(V\) has exactly one basis, up to reordering.
\(\Rightarrow\text{ via contra-positive}:\) Let us assume that \(V\) satisfies neither of these statements, and cover all possible cases of \(V\).
If \(V\) is not finite-dimensional, then we proved in lemma 1 that it doesn’t have exactly one basis.
Otherwise, \(V\) is finite-dimensional, therefore it has a basis (\(2.31\)).
Let \(\mathcal{B}=\left(v_{1},\dots,v_{n}\right)\) be a basis of \(V\).
If \(n=0\), then \(V=\text{Span }\mathcal{B}=\text{Span }\left(\right)=\left\{ 0\right\}\) and it contradicts the fact that \(V\) doesn’t satisfy statement 1.
If \(n=1\) and \(\mathbb{F}\) isn’t isomorphic to \(\mathbb{Z}_{2}\), then we proved in lemma 3 that \(V\) doesn’t have exactly one basis.
If \(n=1\) and \(\mathbb{F}\) is isomorphic to \(\mathbb{Z}_{2}\), then \[V=\text{Span }\left(v_{1}\right)=\left\{ av_{1}\mid a\in\mathbb{F}\right\} =\left\{ 0\cdot v_{1},1\cdot v_{1}\right\} =\left\{ 0,v_{1}\right\}\]
and \(v_{1}\ne0\) because \(\mathcal{B}\) is a basis and is therefore linearly independent - in contradiction to the fact that \(V\) doesn’t satisfy statement 2.
If \(n\ge2\), then we proved in lemma 4 that \(V\) doesn’t have exactly one basis.
This covers all of the possibilities, as required. ∎