Proof.

Notice that it suffices to show that \(S\) is a vector space, or adhere to the then axioms for a vector space. Moreover, A1 and S1 are assumed to be true by the initial condition.

By the second condition, \(\mathbf {0} = 0 u \in S\) and \(-u = (-1) u \in S\). Therefore, A4 and A5 are satisfied. For A2 and A3, \(u, v, w \in V\) is true for \(u, v, w \in S\) by definition. Moreover, because \(V\) is a vector space, A2 and A3 hold. Similarly, because \(V\) is a vector space that satisfies S1, S2, S3, S4, and S5, its subset \(S\) also satisfies the conditions. Therefore, if the two closure conditions are met and \(S\) is a subset of \(V\), then \(S\) is a subspace of \(V\).

Solution.

Notice that \(S\) is a non-empty subset of \(\mathbb {R}^3\), which is a known vector space. Therefore, it suffices to show that \(u = (x_1, 0, z_1), v = (x_2, 0, z_2) \in S\) and scalars \(\alpha \) satisfy the closure conditions for addition and scalar multiplication.

Notice that \(u + v = (x_1 + x_2, 0, z_1 + z_2) \in S\). Moreover, \(\alpha u = (\alpha x_1, 0, \alpha z_1) \in S\). Therefore, \(S\) is a vector space, more specifically a subset of \(\mathbb {R}^3\).

Proof.

First and foremost, the first half of the statement can be proven. If \(\alpha u + \beta v \in S\), then \(u + v \in S\) where \(\alpha = \beta = 1\) since \(\alpha \) and \(\beta \) are all scalars. Similarly, \(\alpha u + \beta v \in S\) implies that \(\alpha u \in S\) where \(\beta = 0\). By Theorem 0.1.2 , \(S\) is a subspace of \(V\).

Continuing with the second half of the statement, if \(S\) is a subset of \(V\), then \(\alpha u, \beta v \in S\). Therefore, \(\alpha u + \beta v \in S\), proving the corollary.

Proof.

First and foremost, notice that \(\mathbf {0} \in \Span (S)\) since all the scalars in linear combination can equal zero. Moreover, let \(u = \sum _{k = 1}^n \alpha _k v_k\) and \(v = \sum _{k = 1}^n \alpha _k v_k\) for any scalars \(\alpha _k\) and \(\beta _k\). By definition, \(\alpha u + \beta v \in \Span (S)\) is true. Therefore, by Corollary 0.1.4 , \(\Span (S)\) is a subspace of \(V\).

Proof.

Let \(u = \sum _{k = 1}^n \alpha _k v_k \in \Span (S)\). By definition, \(v_k \in W\) for integer \(k \in [1, n]\). Moreover, \(u \in W\) by Corollary 0.1.4 and \(\Span (S) \subset W\).

Proof.

First and foremost, notice that by the previous statement, it was shown that \(\Span (S)\) is included in all subspaces that contains \(S\). Let \(A\) be the intersection of all subspaces that contain \(S\). Therefore, \(\Span (S) \subset A\). Moreover, by Theorem 0.1.7 \(\Span (S)\) is a subspace of \(V\) that contains \(S\). Thus, \(A \subset \Span (S)\). Because \(\Span (S) \subset A\) and \(A \subset \Span (S)\), \(\Span (S) = A\).

Proof.

Notice that since all \(t_k \in T\) satisfy \(t_k \in S\), all \(\sum _{k=1}^n c_k t_k \in \Span (T)\) for some appropriate \(n\) and constant \(c_k\) also satisfy \(\sum _{k=1}^n c_k t_k \in \Span (S)\). Therefore, \(\Span (T) \subset \Span (S)\).

Proof.

First and foremost, notice that \(S \subset \Span (S)\). By the first statement, \(\Span (S) \subset \Span (\Span (S))\) holds. Moreover, notice that because \(\Span (S)\) contains all linear combinations of \(S\), making more linear combinations of the linear combinations of \(S\) will not add more elements. Therefore, \(\Span (\Span (S)) \subset \Span (S)\) and \(\Span (\Span (S)) = \Span (S)\).

Proof.

Notice that by the first statement, \(\Span (T) \subset \Span (\Span (S))\). Continuing with the second statement shows that \(\Span (T) \subset \Span (S)\).

Proof.

First, it is self-evident that \(\Span \{ v_1, \ldots , v_k + \lambda v_j, \ldots , v_n \} \subset \Span \{ v_1, \ldots , v_k, \ldots , v_n \}\) since all linear combination of the set of vectors on the left can be represented as the linear combination of the set of vectors on the right. Moreover notice that \(v_k\) can be represented as \(v_k = (v_k + \lambda v_j) - \lambda v_j\) even if \(k = j\). Therefore, \(v_k \in \Span \{ v_1, \ldots , v_k + \lambda v_j, \ldots , v_n \}\) and \(\Span \{ v_1, \ldots , v_k, \ldots , v_n \} \subset \Span \{ v_1, \ldots , v_k + \lambda v_j, \ldots , v_n \}\) by statement 3. Therefore, the equations in the statement hold.

Solution.

By definition, for the set to be linearly independent, all scalars \(c_1\) and \(c_2\) must be zero for the following equation to hold. \[ c_1 \langle 1, 1 \rangle + c_2 \langle 2, 3 \rangle = \langle 0, 0 \rangle \] Writing in system of linear equations, the equation above can be written as the following. \begin{align*} c_1 + 2c_2 &= 0 \\ c_1 + 3c_2 &= 0 \end{align*}

Solving the equation, \(c_1 = c_2 = 0\) and no other set of scalars can lead to the equation. Therefore, the set of vectors is linearly independent.

Proof.

First and foremost, the first half of the theorem can be shown. Consider the set of vectors \(\{ v_1, \ldots , v_{j-1}, v_j, \ldots , v_n \}\). Assume for scalars \(c_k\), the following equation is true. \[ v_j = \sum _{k = 1}^{j-1} c_k v_k \] Therefore, \(\mathbf {0} = \sum _{k = 1}^{j-1} c_k v_k - v_j\). Letting \(c_k = 0\) for integer \(k \in (j, n]\), the following equation holds. \[ \sum _{k = 1}^{j-1} c_k v_k - v_j + \sum _{k = j+1}^{n} c_k v_k = \mathbf {0} \] Because at least one scalar is nonzero, the set of vectors \(\{ v_1, \ldots , v_n \}\) is linearly dependent.

Continuing, let the set of vectors \(\{ v_1, \ldots , v_n \}\) be linearly dependent. For the corresponding scalars \(c_k\), let \(c_j\) be the last scalar that is nonzero. Thus, the following equation holds. \[ \sum _{k = 1}^{j} c_k v_k + \sum _{k = j+1}^{n} 0 v_k = \mathbf {0} \] Rearranging the equation, \(v_k = -\sum _{k = 1}^{j-1} \frac {c_k}{v_j} v_k\), which is a linear combination. Because the converse holds, the theorem is true.

Proof.

By Theorem 0.1.12 , there exists element \(v_j \in S\) such that \(v_j\) is the linear combination of the previous elements as listed in \(S\).

Let \(T = \{ v_1, \ldots , v_{j-1}, v_{j+1}, \ldots , v_n \}\). Consider the following equations for scalars \(d_k\) and \(c_k\). \begin{align*} V &= \sum _{k = 1}^n d_k v_k = \sum _{k = 1}^{j-1} d_k v_k + d_j v_j + \sum _{k = j+1}^n d_k v_k \\ &= \sum _{k = 1}^{j-1} d_k v_k + d_j \left ( \sum _{k = 1}^{j-1} c_k v_k \right ) + \sum _{k = j+1}^n d_k v_k \\ &= \sum _{k = 1}^{j-1} (d_j c_k + d_k) v_k + \sum _{k = j+1}^n d_k v_k = \Span (T) \end{align*}

Therefore, the corollary holds.

Proof.

Let \(S = \{ \mathbf {0}, v_1, \ldots , v_n \}\). Notice that \(\alpha (\mathbf {0}) + \sum _{k = 1}^n 0 v_k = \mathbf {0}\) for any scalar \(\alpha \). Therefore, \(S\) is linearly dependent.

Proof.

Let \(S = \{ v \}\). Notice that if \(v \neq \mathbf {0}\), it is not possible for a nonzero scalar \(\alpha \) to make \(\alpha v = \mathbf {0}\). Therefore, if \(|S| = 1\), then \(v = \mathbf {0}\). For the converse, if \(v = \mathbf {0}\), then any scalar \(\alpha \) will let \(\alpha v = \mathbf {0}\). Therefore the statement holds.

Proof.

Let \(S = \{ v_1, v_2 \}\). First, if \(v_2 = \alpha v_1\) for any scalar \(\alpha \), then \(\mathbf {0} = \alpha v_1 - \alpha v_1 = \alpha v_1 - v_2\). Therefore, \(S\) is linearly dependent. For the converse, \(c_1 v_1 + c_2 v_2 = 0\) for scalars \(c_1\) and \(c_2\). Therefore, assuming that the scalars are nonzero without loss of generality, \(v_2 = -\frac {c_1}{c_2} v_1\). Thus, the statement is true.

Proof.

Let \(S = \{ v_1, \ldots , v_n \}\) and \(T = \{ u_1, \ldots , u_n, v_1, \ldots , v_n \}\). By definition, there exists \(c_k\) that are not all zero such that \(\sum _{k = 1}^n c_k v_k = \mathbf {0}\). Because \(\sum _{k = 1}^n 0 u_k + \sum _{k = 1}^n c_k v_k = \mathbf {0}\), \(T\) is also linearly dependent.

Proof.

Let \(T \subset S\) where \(S\) is linearly independent. Assume for the sake of contradiction that there exists a subset \(T\) that is linearly dependent. However, if \(T\) is linearly dependent, then \(S\) is linearly dependent by the previous property, which is a contradiction. Therefore, by contradiction, if \(S\) is linearly independent then all its subsets \(T\) are also linearly independent.

Proof.

Let \(S = \{ v_1, \ldots , v_n \}\) and \(T = \{ u_1, \ldots , u_m \}\) such that \(S, T \in V\) and \(m > n\). Notice that because \(\Span (S) = V\) and \(T \in V\), there exists a set of constants \(a_{ij}\) that satisfies the following equation (note that we will be using this indices in the next note when we discuss matrices.). \begin{align*} u_1 &= \sum _{j=1}^n a_{1j} v_j \\ u_2 &= \sum _{j=1}^n a_{2j} v_j \\ &\quad \vdots \\ u_m &= \sum _{j=1}^n a_{mj} v_j \end{align*}

Therefore, it suffices to show that there exists constants \(c_1, \ldots , c_m\) such that not all are zero and that satisfies the following equation. \[ \sum _{i=1}^m \sum _{j=1}^n c_i a_{ij} v_j = \mathbf {0} \] Continuing, it suffices to show that the variables \(c_i\) and constants \(a_{ij}\) satisfy the following equations where not all \(c_i\) are zero. \begin{align*} \sum _{i=1}^m c_i a_{i1} &= 0 \\ \sum _{i=1}^m c_i a_{i2} &= 0 \\ &\ \vdots \\ \sum _{i=1}^m c_i a_{in} &= 0 \end{align*}

Notice that there are \(m\) variables and \(n\) equations. It is known that for a linear system of equations, there exists an infinite number of solutions if there are more variables than the equations (don’t worry we will prove this in future notes.). Because \(m > n\), there exists \(c_1, \ldots c_m\) that are not all zero and that satisfy the equations for linear dependence, the theorem holds.

Proof.

By definition, a basis of a vector space \(V\) spans \(V\). Because any linearly dependent subsets of \(V\) contain more than \(n\) elements by Theorem 0.1.20 , any linearly independent subsets of \(V\) must not contain more than \(n\) elements.

Proof.

Let \(S\) and \(T\) be bases of \(V\) with \(p\) and \(q\) elements respectively. By definition, the bases of a vector space are linearly independent. By the previous corollary, it is evident that \(q \leq p\) and \(p \leq q\) are true. Therefore, \(p = q\) and the corollary holds.

Proof.

By definition, a basis of an \(n\)-dimensional vector space \(V\) contains \(n\) elements. By Theorem 0.1.20 , since \(n + 1 > n\), \(S\) must be linearly dependent.

Proof.

For the sake of contradiction, assume that \(u\) can be written as \(u = \sum _{k=1}^n c_k v_k = \sum _{k=1}^n d_k v_k\) where \(c_k \neq d_k\) for some choice of \(k\). Therefore, the following equations can be obtained. \begin{align*} \sum _{k=1}^n c_k v_k &= \sum _{k=1}^n d_k v_k \\ \sum _{k=1}^n (c_k - d_k) v_k &= 0 \end{align*}

Note that by definition, \(S\) is linearly independent and \(c_k - d_k = 0\) for all integer \(k \in [1, n]\). Because it contradicts with the assumption that \(c_k \neq d_k\) for some choice of \(k\), the theorem holds.

Solution.

Consider the following equation. \[ f(x) = 3 \left ( 2x^2 + 1 \right ) - 5 \left ( x^2 \right ) + 2(x) \] Therefore, the coordinate of \(f(x)\) can be represented as the following \(3\)-tuple. \[ \begin {bmatrix} -5 \\ 3 \\ 2 \end {bmatrix}_S \]

Proof.

Notice that by definition, a basis of a vector space \(V\) is a linearly independent set that spans \(V\). Therefore, it suffices to show that there exists a linearly independent set that spans \(V\) that is also a subset of \(S\).

By Theorem 0.1.12 , a subset of \(S\) is linearly dependent if any only if a vector can be represented as the linear combination of the preceding vectors. Let \(T\) be a subset of \(S\) without the vectors that can be represented as the linear combination of the preceding vectors. Notice that \(T\) spans \(V\) as the vector removed can be represented as the linear combination of the other vectors. Moreover, because \(T\) is linearly independent, there exists a basis for \(V\) that is a subset of \(S\).

Proof.

Let \(T\) be a basis for \(V\) and \(A \coloneqq S \cup T\). By definition, \(\Span (A) = V\). Construct \(B \in A\) by removing all vectors that can be represented as a linear combination of the preceding vectors. Notice that no elements from \(S\) will be removed as \(S\) is linearly independent. By construction, \(B\) spans \(V\) and is also linearly independent. Because \(S \subset B\), the statement holds.