OK, I am having a real problem with this and I am desperate.

I have a set of vectors. {(1,0.-1), (2,5,1), (0,-4,3)}

How do I check is this is a basis for $R^3?$

My text says a basis B for a vector space V is a linearly independent subset of V that generates V. OK then. I need to see if these vectors are linearly independent, yes?

If that is so, then for these to be linearly independent the following must be true:

$a_1v_1 + a_2v_2 + ... + a_nv_n $ ≠ 0 for any scalars $a_i$

Is this the case or not?

If it is, then I just have to see if

$a_1(1,0.-1)+ a_2(2,5,1)+ a_3(0,-4,3)$ = 0


$a_1 + 2a_2 + 0a_3 = 0$

$0a_1 + 5a_2 - 4a_3 = 0$

$-a_1 + a_2 + 3a_3 = 0$

has a solution.

Adding these equations up I get $8a_2 - a_3 = 0$ or $a_3 = 8a_2$ so $5a_2 - 32a_2 = 0$ which gets me $a_2 = 0$ and that implies $a_1 = 0 $and$ a_3=0$ as well.

So they are all linearly dependent and thus they are not a basis for $R^3$.

Something tells me that this is wrong. But I am having a hell of a time figuring this stuff out. Please someone help, and I ask: pretend I am the dumbest student you ever met.

  • 3
    $\begingroup$ Instead of saying "it has a solution" you need to say zero is the only solution. So they are linearly independent. Since you have 3 vectors in $R^3$ which are linearly independent then they form a basis. $\endgroup$
    – Maesumi
    Jun 16, 2013 at 2:26

3 Answers 3


A set of vectors $v_1, v_2, ..., v_n$ is linearly independent if and only if we have that

$$a_1v_1 + a_2v_2 + ... + a_nv_n = 0 \;\;$$

only when $ a_1 = a_2 = ... = a_n = 0 $.

(After all, any linear combination of three vectors in $\mathbb R^3$, when each is multiplied by the scalar $0$, is going to be yield the zero vector!)

So you have, in fact, shown linear independence. And any set of three linearly independent vectors in $\mathbb R^3$ spans $\mathbb R^3$. Hence your set of vectors is indeed a basis for $\mathbb R^3$.

  • $\begingroup$ SO that means these vectors ARE a basis, yes? No? $\endgroup$
    – Jesse
    Jun 16, 2013 at 2:19
  • $\begingroup$ Yes it is a basis. $\endgroup$ Jun 16, 2013 at 2:20
  • $\begingroup$ Yes, your set of vectors is a basis for $\mathbb R^3$: they are linearly independent, and they span $\mathbb R^3$ $\endgroup$
    – amWhy
    Jun 16, 2013 at 2:21
  • 1
    $\begingroup$ OK, that's actually a bit of a relief. So this method works to show that a set of vectors, whether it's a single set of points like the ones above or if it was a set of polynomials or whatever, IS a basis for a given space. Yes? $\endgroup$
    – Jesse
    Jun 16, 2013 at 2:23
  • $\begingroup$ Yes, indeed it does! $\endgroup$
    – amWhy
    Jun 16, 2013 at 2:24

Your confusion stems from the fact that you showed that the homogeneous system had only the trivial solution (0,0,0), and indeed homogeneous systems will always have this solution. The criteria for linear dependence is that there exist other, nontrivial solutions.

Another way to check for linear independence is simply to stack the vectors into a square matrix and find its determinant - if it is 0, they are dependent, otherwise they are independent. This method saves a bit of work if you are so inclined.


The easiest way to check whether a given set $\{(a,b,c),(d,e,f),(p,q,r)\} $of three vectors are linearly independent in $\Bbb R^3$ is to find the determinant of the matrix, $$\begin{bmatrix} a & b & c\\ d & e& f\\ p&q&r \end{bmatrix}$$ is zero or not.

If the determinant is zero then the set is linearly dependent else i.e. determinant is nonzero it is linearly independent. Which is the same as saying that the system of linear equations, $$\begin{bmatrix}a&b&c\\d&e&f\\p&q&r\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ has only the trivial solution.


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