Let $V$ be a finite dimensional vector space over $K$ and $A \cup \{ a \} \subseteq V$. Show that $a \in \operatorname{span} A \iff \operatorname{rank} A = \operatorname{rank}\left( A \cup \{ a \}\right)$
My attempt:
We know
$$\operatorname{rank} A = \operatorname{dim}(\operatorname{span}A)$$
And we also know
$$\operatorname{span} A = \{\alpha_1a_1 + \ldots + \alpha_na_n\mid\alpha_1,\ldots,\alpha_n\in K\}.$$
We can write
$$a = \beta_1a_1 + \ldots + \beta_na_n,\ \beta_i\in K.$$
If we do that, then
$$\operatorname{span}\left(A \cup \{ a \}\right) = \{ \alpha_1a_1 + \ldots + \alpha_na_n + a\} = \{ (\beta_1 + \alpha_1)a_1 + \ldots + (\beta_n + \alpha_n)a_n \}$$
We notice that $(\beta_1 + \alpha_1), \ldots, (\beta_n + \alpha_n)$ are just coefficients, we still have the same numbers of $a_i$, that is $a_1, a_2, \ldots, a_n$.
That means the span of $A$ is spanned by the same vectors as $\operatorname{span}\left(A \cup \{ a \}\right)$, so they must have the same dimensions.
If they have the same dimensions, using the equation above $\operatorname{rank} A = \dim(\operatorname{span} A)$, that means we also have
$$\operatorname{rank} A = \dim(\operatorname{span} A) = \dim(\operatorname{span}\left( A \cup \{ a \}\right)) = \operatorname{rank} \left(A \cup \{ a \}\right)$$
Is this proof correct until now ? Or did I miss or forget something ?
If it is written $\iff$, it seems I need to prove it in both directions, how do I do that ?
Best Answer
For the $\Rightarrow$ direction that you did, it's more accurate to write: $$\text{Span} A = \left\{ \alpha_{1}a_{1}+\cdots+\alpha_{n}a_{n}\colon n\geq 1,\text{ }\{\alpha_{i}\}\subset K\text{, }\{a_{i}\}\subset A\right\} $$
since $A$ could be an infinite set, where the span in this case would be the set of every possible finite linear combination. Since this span is a finite dimensional vector space, let $\{a_1, \dots, a_n\}$ be its basis. Then this basis is also a basis for $A \cup \{a\}$, since $a \in \text{Span}A$ and $\{a_1, \dots, a_n\}$ spans $\text{Span}A$. So $\text{Span}(A \cup \{a\})$ has the same basis as $\text{Span}A$, which means they're the same dimension, so $\text{Rank}(A \cup \{a\}) = \text{Rank}A$.
For the other direction, suppose that $\text{Rank}(A \cup \{a\}) = \text{Rank}A$. Let $\text{Rank}A = n$ and let $\{a_1, \dots, a_n\}$ be a basis for $\text{Span}A$. Then $\{a_1, \dots, a_n, a\}$ is a spanning set for $\text{Span}(A \cup \{a\})$, but $\text{Span}(A \cup \{a\})$ has the same dimension as $\text{Span}A$, so $\{a_1, \dots, a_n, a\}$ couldn't be linearly independent otherwise it'd be a basis of length $n+1$ which would mean $\text{Rank}(A \cup \{a\})=n+1 > n$. So the set is linearly dependent, i.e. there exists $\{\alpha_i\} \subset K$ such that $\alpha_1 a_1 + \dots + \alpha_n a_n + \alpha_{n+1} a = 0$ where not all $\alpha_i = 0$. If $a = 0$ then $a \in \text{Span}A$ anyway, so suppose that $a \neq 0$. Then $\alpha_{n+1} \neq 0 $ too, otherwise $\alpha_1 a_1 + \dots + \alpha_n a_n = 0$ is satisfied with not all $\alpha_i=0$, but $\{\alpha_1, \dots, \alpha_n\}$ is a basis so is linearly independent. Therefore we can write $a = \alpha_{n+1}^{-1} \alpha_1 a_1 + \dots + \alpha_{n+1}^{-1}\alpha_n a_n$ so $a \in \text{Span}A$.