# Proof that union of linearly independent set with element not in the span of the set is linearly independent

I need to proof the following theorem and I would appreciate if you could check if my proof is sound. I'm generally new to proofs, so any recommendations on proof strategy/style are more than welcome. Thank you.

$$\textbf{Theorem:}$$

Let $$S=\{\textbf{v}_1, \textbf{v}_2, ..., \textbf{v}_k\}$$ be a linearly independent subset of some vector space $$V$$. Let $$\textbf{u} \in V$$. If $$\textbf{u} \notin \operatorname{span}(S)$$, then $$S \cup \{\textbf{u}\}$$ is linearly independent.

$$\textbf{Proof:}$$ I use a contrapositive proof, i.e. $$\lnot Q \Rightarrow \lnot P$$. Suppose $$S \cup \{ \textbf{u} \}$$ is linearly $$\textbf{dependent}.$$ Then, by definition of linear dependence, there exist $$\alpha_1, \alpha_2, ..., \alpha_{k+1} \in \mathbb{R}$$ not all zero s.t.

$$\sum_{i=1}^{k+1}\alpha_i v_i = 0$$ $$= \sum_{i=1}^{k}\alpha_i v_i + \alpha_j u = 0$$ where, without loss of generality, $$\alpha_j u= \alpha_{k+1}v_{k+1}$$.

Then it follows, $$\alpha_j u = -\sum_{i=1}^{k}\alpha_i v_i$$ $$= u =\sum_{i=1}^{k}\alpha_i^{'} v_i$$ where $$\alpha_i^{'} = \frac{-\alpha_i}{\alpha_j}$$, for $$\alpha_j \neq 0$$, implying that $$u \in \operatorname{span}(S) \cup \{u\}$$.

Thus, we can conclude that if $$u \notin \operatorname{span}(S)$$, $$S \cup \{u\}$$ is linearly independent. This completes the proof.

• A clear way to know that your proof is incomplete is that you haven’t used the linear independence of $S$ supposistion in your proof. The good news is, other than some weird notational quirks, your proof is almost complete. You have divided by $\alpha_j$ stating that $\alpha_j$ is non-zero, but younhavent justifies why it is non-zero. Also, you say at the end that $u$ is in Span$(S) \cup \{u\}$, but this is true for every $S$, so I’m unsure what you mean here? – Adam Higgins Dec 27 '18 at 12:15

Suppose$$S\cup\{\mathbf{u}\}$$ is linearly dependent. Then there exist coefficients $$\alpha_1,\dots,\alpha_k\:$$ and $$\:\beta$$, not all $$0$$, such that $$\sum_{i=1}^k\alpha_i\,\mathbf{v}_i+\beta\, \mathbf{u}=0.$$ Observe that $$\beta\ne 0$$, for otherwise, $$S$$ would be linearly dependent, contrary to the hypotheses. So we can deduce that $$\mathbf u=\sum_{i=1}^k\Bigl(-\tfrac{\alpha_i}\beta\Bigr)\mathbf{v}_i,$$ which means$$\;\mathbf u\in\operatorname{span} S$$.