Is this set of vectors a basis?

Question

I was asked to tell if the following vectors-set is a basis (vector space not menionted), so after reduction I got the following: $$\left( {\begin{array}{*{20}{c}} 0 & 1 & 1 \\ 2 & 1 & 3 \\ { - 1} & 1 & 0 \\ \end{array}} \right) = \left( \begin{array}{l} x \\ y \\ z \\ \end{array} \right) \Rightarrow \left( {\begin{array}{*{20}{c}} { - 1} & 1 & 0 \\ 0 & 3 & 3 \\ 0 & 0 & 0 \\ \end{array}} \right) = \left( \begin{array}{l} z \\ y + 2z \\ x - \frac{y}{3} - \frac{{2z}}{3} \\ \end{array} \right)$$

So, at this point in order to prove it is not a basis, I need to find $x,y,z$ such that are not linear combination of the three vectors. Right?
How do you that efficiently, just by trial and error?

The notation is pretty awkward here; a $3 \times 3$-matrix can never equal a $3 \times 1$-vector. — fuglede, Jan 04 '14 at 12:58
You right. I wasn't sure how to right it properly using latex — AndrePoole, Jan 04 '14 at 13:23

Adriano · Accepted Answer · 2014-01-04T13:03:09.153

1

Notice that the third column vector is the sum of the first two column vectors. Hence, the three column vectors are linearly dependent, so they do not form a basis of $\mathbb R^3$.

If you didn't notice this, then you can reduce the matrix of column vectors into row echelon form like you did. Because you obtained a row of zeroes, the vectors must be linearly dependent.

edited Jan 04 '14 at 13:03

answered Jan 04 '14 at 12:57

Adriano

41,576

OK, nice one. it's a better approach – AndrePoole Jan 04 '14 at 13:02
Why "Because you obtained a row of zeroes, the vectors must be linearly dependent" is true? – AndrePoole Jan 04 '14 at 13:24
@AndrePoole: Do you know the terminology of row and column ranks? That the vectors are linearly dependent is the same as saying that the column rank is strictly less than $3$. Now, column rank equals row rank, and does not change under elementary row operations. Since the row rank of your reduced matrix is clearly less than $3$, so is the column rank of the matrix you began with. – fuglede Jan 04 '14 at 16:54

Is this set of vectors a basis?

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