Let's say set $S = \{(2,1,4), (1,-1,1), (3,2,5)\}$ and the vector space V is $R^3$. In this case, the number of independent vectors is equal to the dimension of the vector space. I know that one way to show that S is a spanning set for V is by showing that a linear combination of the vectors can equal any vector $(a,b,c)$. My professor told us to set up an augmented matrix and show that the rank of A is equal to the rank of A|b.

I would like to know if there is a less tedious method. If I should stick to this method, I would like to know why.


Let's say $S = \{(2,1,4), (1,-1,1), (3,2,5), (1,4,-2) \}$

How can I prove that a set spans a vector space if the number of independent vectors is larger than the dimension of the vector space? I won't be able to show that the set is a basis, and therefore also a spanning set. Is there a quick proof?

Edit 2 (for Future Reference)

The set S cannot have 4 independent vectors because the vectors are in $R^3$. If set S has more than 3 vectors it can still span $R^3$. That can be proven by finding a basis within the set.

  • $\begingroup$ I'm not clear what you want to show. If you want to show that $S$ spans $V$, then you're done as $|S| = 3 = \dim R^3$. Prove that if $S$ is a linearly indep. set of the same size as the dimension of the space, then the set must span the space. $\endgroup$ Apr 8, 2014 at 22:08
  • $\begingroup$ What does |S| mean? $\endgroup$
    – Tad
    Apr 8, 2014 at 22:15

1 Answer 1


The determinant test is worthwhile. Let $M$ be a matrix whose column vectors are from $S$. The vectors in $S$ form a basis if and only if $det(M) \neq 0$.

I figure I'll edit my answer so nobody has to dig through all the comments (and in light of the edit on the OP). A basis is a maximally independent set. It also spans the space. So if $|S| > dim(V)$, then it is necessary to find a basis to show $S$ spans $V$. There are $\binom{|S|}{dim(V)}$ basis candidates to choose from $S$. The general way to construct a basis is to start with an empty set and add in vectors. If the vector you are examining makes the set linearly dependent, throw it out. The multiples test is one test. If you see $v = kx$, for some $x$ in your set and $k \in \mathbb{F}$, then $v$ and $x$ are linearly dependent. You could also row-reduce to determine linear (in)dependence. Once you can construct a square matrix, you can use the determinant test.

  • $\begingroup$ But the determinant test shows that the vectors from S are independent. I want to show that the set spans V. $\endgroup$
    – Tad
    Apr 8, 2014 at 22:12
  • $\begingroup$ The dimension of $\mathbb{R}^{3}$ is $3$. So if you have three independent vectors in $\mathbb{R}^{3}$, you have a basis. $\endgroup$
    – ml0105
    Apr 8, 2014 at 22:13
  • $\begingroup$ Let's say there were 4 independent vectors in S. Can I automatically assume S spans $R^3$? $\endgroup$
    – Tad
    Apr 8, 2014 at 22:17
  • $\begingroup$ Yes, because any three of the vectors form a basis for $\mathbb{R}^{3}$. A basis is a minimum spanning set. $\endgroup$
    – ml0105
    Apr 8, 2014 at 22:21
  • 1
    $\begingroup$ You can never have more than three independent vectors in $\mathbb R^3$ $\endgroup$ Apr 8, 2014 at 22:27

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