Proof. Let \(\mathcal{T}\) be a finite subset of \(\mathcal{S}\cup\{v_0\}.\) If \(v_0 \notin \mathcal{T},\) then \(\mathcal{T}\) is linearly independent, as \(\mathcal{S}\) is linearly independent. So suppose \(v_0 \in \mathcal{T}.\) There exist distinct elements \(v_1,\ldots,v_n\) of \(\mathcal{S}\) so that \(\mathcal{T}=\{v_0,v_1,\ldots,v_n\}.\) Suppose \(s_0v_0+s_1v_1+\cdots+s_nv_n=0_V\) for some scalars \(s_0,s_1,\ldots,s_n \in \mathbb{K}.\) If \(s_0 \neq 0,\) then we can write \[v_0=-\sum_{i=1}^n \frac{s_i}{s_0}v_i,\] contradicting the assumption that \(v_0 \notin \operatorname{span}(\mathcal{S}).\) Hence we must have \(s_0=0.\) Since \(s_0=0\) it follows that \(s_1v_1+\cdots+s_nv_n=0_V\) so that \(s_1=\cdots=s_n=0\) by the linear independence of \(\mathcal{S}.\) We conclude that \(\mathcal{S}\cup\{v_0\}\) is linearly independent.

Proof. Since \(v_0 \in \operatorname{span}(\mathcal{S} \setminus\{v_0\}),\) there exist vectors \(v_1,\ldots,v_n \in \mathcal{S}\) with \(v_i\neq v_0\) and scalars \(s_1,\ldots,s_n\) so that \(v_0=s_1v_1+\cdots+s_n v_n.\) Suppose \(v \in V.\) Since \(\mathcal{S}\) is a generating set, there exist vectors \(w_1,\ldots,w_k \in \mathcal{S}\) and scalars \(t_1,\ldots,t_k\) so that \(v=t_1w_1+\cdots+t_kw_k.\) If \(\{w_1,\ldots,w_k\}\) does not contain \(v_0,\) then \(v \in \operatorname{span}(\mathcal{S}\setminus\{v_0\}),\) so assume that \(v_0 \in \{w_1,\ldots,w_k\}.\) After possibly relabelling the elements of \(\{w_1,\ldots,w_k\}\) we can assume that \(v_0=w_1.\) Hence we have \[v=t_1\left(s_1v_1+\cdots+s_n v_n\right)+t_2w_2+\cdots+t_kw_k\] with \(v_0 \neq v_i\) for \(1\leqslant i\leqslant n\) and \(v_0 \neq w_j\) for \(2\leqslant j\leqslant k.\) It follows that \(v \in \operatorname{span}(\mathcal{S}\setminus\{v_0\}),\) as claimed.

Proof. We show that if \(\mathcal{T}\) has exactly \(n+1\) elements, then it is linearly dependent. In the other cases, \(\mathcal{T}\) contains a subset with exactly \(n+1\) elements and if this subset is linearly dependent, then so is \(\mathcal{T}.\)

We prove the claim by induction on \(n\geqslant 0.\) Let \(\mathcal{A }(n)\) be the following statement: “For any \(\mathbb{K}\)-vector space \(V,\) if there exists a generating subset \(\mathcal{S}\subset V\) with \(n\) elements, then all subsets of \(V\) with \(n+1\) elements are linearly dependent.”

We first show that \(\mathcal{A }(0)\) is true. A subset with zero elements is the empty set \(\emptyset.\) Hence \(V=\operatorname{span}(\emptyset)=\{0_V\}\) is the zero vector space. The only subset of \(\{V\}\) with \(1\) element is \(\{0_V\}.\) Since \(s0_V=0_V\) for all \(s\in \mathbb{K}\) , the set \(\{0_V\}\) is linearly dependent, thus showing that \(\mathcal{A }(0)\) is correct.

Suppose \(n\geqslant 1\) and that \(\mathcal{A }(n-1)\) is true. We want to argue that \(\mathcal{A }(n)\) is true as well. Suppose \(V\) is generated by the set \(\mathcal{S}=\{v_1,\ldots,v_n\}\) with \(n\) elements. Let \(\mathcal{T}=\{w_1,\ldots,w_{n+1}\}\) be a subset with \(n+1\) elements. We need to show that \(\mathcal{T}\) is linearly dependent. Since \(\mathcal{S}\) is generating, we have scalars \(s_{ij} \in \mathbb{K}\) with \(1\leqslant i\leqslant n+1\) and \(1\leqslant j\leqslant n\) so that \[\tag{3.10} w_i=\sum_{j=1}^n s_{ij}v_j\] for all \(1\leqslant i\leqslant n+1.\) We now consider two cases:

Case 1. If \(s_{11}=\cdots=s_{{n+1},1}=0,\) then (3.10) gives for all \(1\leqslant i\leqslant n+1\) \[w_i=\sum_{j=2}^ns_{ij}v_j.\] Notice that the summation now starts at \(j=2.\) This implies that \(\mathcal{T}\subset W,\) where \(W=\operatorname{span}\{v_2,\ldots,v_n\}.\) We can now apply \(\mathcal{A }(n-1)\) to the vector space \(W,\) the generating set \(\mathcal{S}_1=\{v_2,\ldots,v_n\}\) and the subset with \(n\) elements being \(\mathcal{T}_1=\{w_1,\ldots,w_n\}.\) It follows that \(\mathcal{T}_1\) is linearly dependent and hence so is \(\mathcal{T},\) as it contains \(\mathcal{T}_1.\)

Case 2. Suppose there exists \(i\) so that \(s_{i1}\neq 0.\) Then, after possibly relabelling the vectors, we can assume that \(s_{11}\neq 0.\) For \(2\leqslant i\leqslant n+1\) we thus obtain from (3.10) \[\begin{aligned} w_i-\frac{s_{i1}}{s_{11}}w_1&=w_i-\frac{s_{i1}}{s_{11}}\left(\sum_{j=1}^ns_{1j}v_j\right)=\sum_{j=1}^ns_{ij}v_j-\frac{s_{i1}}{s_{11}}\left(\sum_{j=1}^ns_{1j}v_j\right)\\ &=\sum_{j=1}^n\left(s_{ij}-\frac{s_{i1}}{s_{11}}s_{1j}\right)v_j\\ &=\underbrace{\left(s_{i1}-\frac{s_{i1}}{s_{11}}s_{11}\right)}_{=0}v_1+\sum_{j=2}^n\left(s_{ij}-\frac{s_{i1}}{s_{11}}s_{1j}\right)v_j\\ &=\sum_{j=2}^n\left(s_{ij}-\frac{s_{i1}}{s_{11}}s_{1j}\right)v_j. \end{aligned}\] Hence, setting \[\tag{3.11} \hat{w}_i=w_i-\frac{s_{i1}}{s_{11}}w_1\] for \(2\leqslant i\leqslant n+1\) and \(\hat{s}_{ij}=s_{ij}-\frac{s_{i1}}{s_{11}}s_{1j}\) for \(2\leqslant i\leqslant n+1\) and \(2\leqslant j\leqslant n,\) we obtain the relations \[\hat{w}_i=\sum_{j=2}^n\hat{s}_{ij}v_j\] for all \(2\leqslant i\leqslant n+1.\) Therefore, the set \(\hat{\mathcal{T}}=\{\hat{w}_2,\ldots,\hat{w}_{n+1}\}\) with \(n\) elements is contained in \(W\) which is generated by \(n-1\) elements. Applying \(\mathcal{A }(n-1),\) we conclude that \(\hat{\mathcal{T}}\) is linearly dependent. It follows that we have scalars \(t_2,\ldots,t_{n+1}\) not all zero so that \[t_2\hat{w}_2+\cdots+t_{n+1}\hat{w}_{n+1}=0_V.\] Using (3.11), we get \[\sum_{i=2}^{n+1}t_i\left(w_i-\frac{s_{i1}}{s_{11}}w_1\right)=-\left(\sum_{i=2}^{n+1}t_i\frac{s_{i1}}{s_{11}}\right)w_1+t_2w_2+\cdots+t_{n+1}w_{n+1}=0_V.\] Since not all scalars \(t_2,\ldots,t_{n+1}\) are zero, it follows that \(w_1,\ldots,w_{n+1}\) are linearly dependent and hence so is \(\mathcal{T}.\)

(iv) is an immediate consequence of (iii).

Proof. Let \(V\) be a finite dimensional \(\mathbb{K}\)-vector space and \(U\subset V\) a subspace. Let \(\mathcal{S}=\{v_1,\ldots,v_n\}\) be a basis of \(V.\) For \(1\leqslant i\leqslant n,\) we define \(U_i=U\cap\operatorname{span}\{v_1,\ldots,v_i\}.\) By construction, each \(U_i\) is a subspace and \(U_1\subset U_2 \subset \cdots\subset U_n=U,\) since \(\mathcal{S}\) is a basis of \(V.\)

We will show inductively that all \(U_i\) are finite dimensional. Notice that \(U_1\) is a subspace of \(\operatorname{span}\{v_1\}.\) The only subspaces of \(\operatorname{span}\{v_1\}\) are \(\{0_V\}\) and \(\{tv_1\,|\,t \in \mathbb{R}\},\) both are finite dimensional, hence \(U_1\) is finite dimensional.

Assume \(i\geqslant 2.\) We will show next that if \(U_{i-1}\) is finite dimensional, then so is \(U_{i}.\) Let \(\mathcal{T}_{i-1}\) be a basis of \(U_{i-1}.\) If \(U_i=U_{i-1},\) then \(U_i\) is finite dimensional as well, so assume there exists a non-zero vector \(w \in U_{i}\setminus U_{i-1}.\) Since \(\mathcal{S}\) is a basis of \(V\) and since \(w \in \operatorname{span}\{v_1,\ldots,v_i\},\) there exist scalars \(s_1,\ldots,s_i\) so that \(w=s_1v_1+\cdots+s_iv_i.\) By assumption, \(w \notin U_{i-1},\) hence \(s_i\neq 0.\) Any vector \(v \in U_i\) can be written as \(v=t_1v_1+\cdots+t_iv_i\) for scalars \(t_1,\ldots,t_i.\) We now compute \[\begin{aligned} v-\frac{t_i}{s_i}w&=\sum_{k=1}^it_kv_k-\frac{t_i}{s_i}\left(\sum_{k=1}^is_kv_k\right)=\sum_{k=1}^i\left(t_k-\frac{t_i}{s_i}s_k\right)v_k\\ &=\sum_{k=1}^{i-1}\left(t_k-\frac{t_i}{s_i}s_k\right)v_k \end{aligned}\] so that \(v-(t_i/s_i)w\) can be written as a linear combination of the vectors \(v_1,\ldots,v_{i-1},\) hence is an element of \(U_{i-1}.\) Recall that \(\mathcal{T}_{i-1}\) is a basis of \(U_{i-1},\) hence \(v-(t_i/s_i)w\) is a linear combination of elements of \(\mathcal{T}_{i-1}.\) It follows that any vector \(v \in U_i\) is a linear combination of elements of \(\mathcal{T}_{i-1}\cup \{w\},\) that is, \(\mathcal{T}_{i-1}\cup \{w\}\) generates \(U_i.\) Since \(\mathcal{T}_{i-1}\cup\{w\}\) contains finitely many vectors, it follows that \(U_i\) is finite dimensional.

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.

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