Proof that basis is linearly generating set of some subspace $V$.

linear algebraproof-explanationproof-verification

Good morning.

I've been preparing for my linear algebra exam and went through a proof we did in class by our professor. I've been looking at it for a while and I can't quite understand why the proof below is correct. I would really appreciate if someone could explain to me why this proof works.

In particular, I am confused about one thing:

  • In Part 1 of the proof, the professor notes that we have to show that $B$ is a linearly independent set (which seems obvious). However, we work towards and arrive at the the conclusion that $B$ without $\textbf{v}_1$ is still a generating set, thus "proving" the first part. I don't quite get how the proof shows that $B$ is linearly independent. Am I missing something? See the Theorem and proof below. Thanks.

$\textbf{Theorem}$:

Let $B$ be a subset of some vector space $V$ . Then the following are equivalent:

  1. $B$ is a basis for $V$.
  2. $B$ is linearly independent generating set of $V$.

$\textbf{Proof}$

Part 1: $1. \Rightarrow 2.$

Assume that the subset $B \subset V$ is a basis for $V$. Then, by definition, $B$ is a minimal generating subset of $V$, where it follows by definition that $\operatorname{span}(B)=V$.

Indirect proof:

We have to show that $B$ is linearly independent.

Assume that $B$ is linearly dependent. Then there must exist $\alpha_i \in \mathbb{R}$ not all zero such that
$$\sum_{i=1}^{k} \alpha_i \textbf{v}_i = 0$$
Then, without loss of generality,
$$\sum_{i=2}^{k} \alpha_i \textbf{v}_i = \textbf{v}_1$$
We now show that $B \setminus \{\textbf{v}_1\}$ is still a generating set.

Let $\textbf{u} \in V$ be some arbitrary element. Since $B$ is a generating set, there exist $\beta_i \in \mathbb{R}$ such that
$$\sum_{i=1}^{k} \beta_i \textbf{v}_i = \textbf{u}$$
Then,
$$\sum_{i=2}^{k} \beta_i \textbf{v}_i + \beta_1 \textbf{v}_1= \textbf{u}$$
$$= \sum_{i=2}^{k} \beta_i \textbf{v}_i + \beta_1 \sum_{i=2}^{k} \alpha_i \textbf{v}_i = \textbf{u} $$
$$= \sum_{i=2}^{k} (\beta_1 \alpha_i + \beta_i) \textbf{v}_i \in \operatorname{span}(B)$$
Since $\textbf{u}$ was arbitrary,
$$\operatorname{span}(B \setminus \{ \textbf{v}_1 \} )= V$$
Therefore, we conclude that $B \setminus \{ \textbf{v}_1 \}$ is a generating set.

Part 2: $1. \Leftarrow 2.$
Assume that $B$ is a linearly independent generating set of $V$. We have to show that $B$ is a basis for $V$.
Suppose that, without loss of generality, that $B \setminus \{\textbf{v}_1 \}$ is still a generating set. Then it follows that
$$\sum_{i=2}^{k} \alpha_i \textbf{v}_i = \textbf{v}_1$$
for some $\alpha_i \in \mathbb{R}$. This implies that $B$ is not linearly independent, a contradiction.

Best Answer

Your professor defined “basis” as a minimal generating subset of your vector space. Now, the goal is to prove that if $B$ is a basis of $V$, then $B$ is linearly independent and spans $V$. It is trivial that it spans $V$ since, by definition, $B$ is a generating set. So, all that remains to be proved is that $B$ is linearly independent. Your professor proves this by contridiction: if $B$ was not linearly independent, then there would be a non-trivial linear combination of its elements whose sum would be $0$. But then, some element $v\in B$ would be a linear combination of the remaining elements of $B$. So, if you remove $v$ from $B$, then you still have a generating set. This is impossible, since you are assuming that $B$ is a minimal generating set and $B\setminus\{v\}\varsubsetneq B$.

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