📂Linear Algebra

Basis Addition/Subtraction Theorem

Theorem¹

Let $S$ be a non-empty subset of vector space $V$.

(a) If $S$ is linearly independent and if $\mathbf{v} \in V$ equals $\mathbf{v} \notin \text{span}(S)$, then $S \cup \left\{ \mathbf{v} \right\}$ remains linearly independent.

(b) If $\mathbf{v} \in S$ can be represented as a linear combination of other vectors in $S$, then $S$ and $S \setminus \left\{ \mathbf{v} \right\}$ span the same space. That is, the following holds:

$$\text{span}(S) = \text{span} \left( S \setminus \left\{ \mathbf{v} \right\} \right)$$

Proof

(a)

Assume that $S=\left\{ \mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{r} \right\} \subset V$ is linearly independent and that $\mathbf{v} \notin \text{span}(S)$. To show that $S \cup \left\{ \mathbf{v} \right\}$ is linearly independent, it suffices to demonstrate that the only solution to the following equation

$$\begin{equation} k_{1} \mathbf{v}_{1} + k_{2} \mathbf{v}_{2} + \cdots + k_{r} \mathbf{v}_{r} + k \mathbf{v} = \mathbf{0} \label{eq1} \end{equation}$$

is $k_{1}=k_{2}=k_{r}=k=0$.

However, if $k \ne 0$, then $k \mathbf{v} = -\sum \limits_{i=1}^{r} k_{i} \mathbf{v}_{i} \in \text{span} (S)$ holds. Therefore, $k=0$ must be true, leading to the following equation:

$$k_{1} \mathbf{v}_{1} + k_{2} \mathbf{v}_{2} + \cdots + k_{r} \mathbf{v}_{r} = \mathbf{0}$$

Given the assumption that $S$ is linearly independent, the only solution to this equation is $k_{1}=k_{2}=k_{r}=0$. Thus, the only solution for $(1)$ is $k_{1}=k_{2}=k_{r}=k=0$, proving that $S \cup \left\{ \mathbf{v} \right\}$ is linearly independent.

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(b)

Assuming $S=\left\{ \mathbf{v}_{1}, \mathbf{v}_{2}, \dots, \mathbf{v}_{r} \right\} \subset V$ and that $\mathbf{v}_{r}$ can be expressed as a linear combination of other vectors as follows:

$$\begin{equation} \mathbf{v}_{r} = c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + \cdots +c_{r-1}\mathbf{v}_{r-1} \label{eq2} \end{equation}$$

Now, let $\mathbf{w} \in \text{span} (S)$. According to the definition of spanning, $\mathbf{w}$ can be expressed as:

$$\mathbf{w} = k_{1} \mathbf{v}_{1} + k_{2} \mathbf{v}_{2} + \cdots + k_{r-1} \mathbf{v}_{r-1} + k_{r} \mathbf{v}_{r}$$

Substituting $(2)$ into the equation above results in:

$$\begin{align*} \mathbf{w} &= k_{1} \mathbf{v}_{1} + k_{2} \mathbf{v}_{2} + \cdots + k_{r-1} \mathbf{v}_{r-1} + k_{r} \mathbf{v}_{r} \\ &= k_{1} \mathbf{v}_{1} + k_{2} \mathbf{v}_{2} + \cdots + k_{r-1} \mathbf{v}_{r-1} + k_{r} \left( c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + \cdots +c_{r-1}\mathbf{v}_{r-1}\right) \\ &= \left( k_{1} + k_{r}c_{1} \right)\mathbf{v}_{1} + \left( k_{2} + k_{r}c_{2} \right) \mathbf{v}_{2} + \cdots + \left( k_{r-1} + k_{r}c_{r-1} \right) \mathbf{v}_{r-1} \end{align*}$$

Therefore, $\mathbf{w} \in \text{span} (S \setminus \left\{ \mathbf{v} \right\})$ is true. The converse is also true by the same logic, hence the two sets are identical.

$$\text{span}(S) = \text{span} \left( S \setminus \left\{ \mathbf{v} \right\} \right)$$

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