# Set of $n$ vectors spanning an $n$ - dim vector space implies set is a basis

Theorem:

If a set of n vectors spans an n-dimensional vector space, then the set is a basis for that vector space.

Attempt:

Let S be a set of n vectors spanning an n-dimensional vector space. This implies that any vector in the vector space $$\left(V, R^{n}\right)$$ is a linear combination of vectors in the set S.

It suffice to show that S is linearly independent.

Suppose that S is linearly dependent. Then for some vectors $$\vec{v}_{i}, \exists i \in \mathbb{Z}_{1}^{n}$$ in S that may be expressed as a linear combination of some vectors in S, the removal of $$\vec{v}_{i}$$ does not affect the span of set S.

Any hints to bring me forward is highly appreciated.

Thanks in advance.

• You're almost done - say $span(S) = span(S \setminus \lbrace v_1 \rbrace)$, then $n = \dim(span(S)) = \dim(span(S \setminus \lbrace v_1 \rbrace))$, but what is $|S \setminus \lbrace v_1 \rbrace|$? Nov 23, 2018 at 14:34
• @Stockfish And so a contradiction exists for the order of the set S is n is n -1. Nov 23, 2018 at 14:40

## 2 Answers

Let $$S=\{x_1,x_2,\ldots,x_n\}$$ be a set spanning the vector space $$V$$ of dimension $$n$$. Suppose $$S$$ is not a basis of $$V$$. Then $$S$$ is linearly dependent. Thus there exists $$x_i\in S$$ such that $$x_i$$ is a linear combination of remaining $$n-1$$ vectors as $$S\setminus \{x_i\}$$ also spans $$V$$. Thus $$n-1$$ vectors can span $$V$$, which is a contradiction. Hence $$S$$ is a basis.

• Which is a contradiction or not, depending on what previous results we're allowed to use... Nov 23, 2018 at 14:39
• Up for the effort Nov 23, 2018 at 14:41
• "Thus n−1 vectors can span V, which is a contradiction" why is this a contradiction? Oct 6, 2021 at 20:56

It's impossible to say what's a correct solution without knowing what previously proved results you're allowed to use (for example, the other answer is fine if you're given that $$n-1$$ vectors cannot span $$V$$.)

Since the word "dimension" appears it seems reasonable to conjecture that you've already shown that any basis for $$V$$ must have exactly $$n$$ elements; this is required before the typical definition of "dimension" makes sense. Assuming that it's an easy exercise:

Show that $$S$$ contains a minimal spanning set: There exists $$S_1\subset S$$ such that $$S_1$$ spans $$V$$ and if $$S_2\subset S_1$$ and $$S_2$$ spans $$V$$ then $$S_2=S_1$$. Show that it follows that $$S_1$$ is independent.

So $$S_1$$ is a basis. Hence that previous result show that $$S_1$$ contains exactly $$n$$ elements; since $$S_1\subset S$$ this shows that $$S_1=S$$.