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


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.

  • 1
    $\begingroup$ 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 $|S \setminus \lbrace v_1 \rbrace|$? $\endgroup$
    – Stockfish
    Nov 23, 2018 at 14:34
  • $\begingroup$ @Stockfish And so a contradiction exists for the order of the set S is n is n -1. $\endgroup$ Nov 23, 2018 at 14:40

2 Answers 2


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.

  • 4
    $\begingroup$ Which is a contradiction or not, depending on what previous results we're allowed to use... $\endgroup$ Nov 23, 2018 at 14:39
  • $\begingroup$ Up for the effort $\endgroup$ Nov 23, 2018 at 14:41
  • $\begingroup$ "Thus n−1 vectors can span V, which is a contradiction" why is this a contradiction? $\endgroup$ 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 $S_1=S$.


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