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Explain why $S$ is not a basis for $\mathbb{R}^3$

$S=\{(1, 3, 0),(4, 1, 2),(-2, 5, -2)\}$

I set this equal to an arbitrary vector $\mathbf{x} = (x_1, x_2, x_3)$

After solving I got the matrix:

$\begin{bmatrix} 1 & 4 & -2 \\ 3 & 1 & 5 \\ 0 & 2 & -2\end{bmatrix}$

And since the $det(A)=0$ this set does not span $\mathbb{R}^3$. However, I don't understand the answer the solution gave. It says:

A basis for $\mathbb{R}^3$ contains three linearly independent vectors. Because $-2(1, 3, 0)+(4, 1, 2)+(-2, 5, -2)=(0, 0, 0)$

$S$ is linearly dependent and is, therefore, not a basis for $\mathbb{R}^3$

However, if I test for linear independence I get the following $rref(A)$:

$\begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

Which would imply to me that $S$ is linearly independent given the $0$ column.

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Your RREF tells you that $S$ is linearly independent. What you have done is row-reduced the augmented matrix: $$\left(\begin{array}{ccc|c} 1 & 4 & -2 & 0\\ 3 & 1 & 5 & 0\\ 0 & 2 & -2 & 0 \end{array}\right)$$ to the reduced row echelon form $$\left(\begin{array}{ccc|c} 1 & 0 & 2 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 0 & 0 \end{array}\right)$$ What this tells you is that the matrix formed by setting the columns equal to the given vectors of $S$ has non-trivial nullspace, meaning there exist constants $c_1,c_2,c_3$ not all $0$, such that $c_1 (1,3,0) + c_2(4,1,2) + c_3(-2,5,-2) = (0,0,0)$

What we will have done is effectively checked whether the 3 vectors given satisfy the definition of linear independence. Since we have nonzero constants $c_1,c_2,c_3$ so that $c_1 (1,3,0) + c_2(4,1,2) + c_3(-2,5,-2) = (0,0,0)$, they are not linearly independent.

In fact, by simply backsolving and picking $c_3 = 1$, we could choose $c_2 = 1$, $c_1 = -2$

That implies that $\{(1,3,0),(4,1,2),(-2,5,-2)\}$ are linearly dependent, as the solution in the book states. And 3 vectors in $\mathbb{R}^3$ are a basis of $\mathbb{R}^3$ if and only if they are linearly independent, so the 3 vectors given are not a basis of $\mathbb{R}^3$.


I previously said that this is because you have a row of zeros. Technically, this is wrong. The correct explanation has to do with pivot columns, namely that we have a non-leading non-pivot column on the LHS when reducing to row echelon form. A row of zeros is enough if you are dealing with a square matrix, but seeing as that is rarely the case, you will want to know the more general method.

Say you have $4$ vectors in $\mathbb{R}^6$. These are linearly independent if and only if the matrix formed with the vectors has rank $4$. Another way to verify this is that if each and every column on the LHS of the augmented matrix (when reduced to row echelon form) is a pivot column, then the column vectors forming your matrix are linearly independent.

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  • $\begingroup$ Thank you for this explanation! This is what I was looking for. $\endgroup$ – hax0r_n_code Apr 23 '14 at 1:53
  • $\begingroup$ You are quite welcome. I went ahead and elaborated a bit on what the mechanics are behind seeing whether 3 vectors in $\mathbb{R}^3$, or for that matter, any set of $n$ vectors in $\mathbb{R}^n$, is a basis of $\mathbb{R}^n$ (equivalently, linearly independent). In practice, set the vectors up as columns in a matrix, and row-reduce to compute the nullspace of the matrix. If the nullspace is nontrivial (i.e., if the nullspace contains a nonzero vector), then the vectors are linearly dependent, and cannot be a basis of $\mathbb{R}^n$. $\endgroup$ – Nicholas Stull Apr 23 '14 at 1:59
  • $\begingroup$ Strictly speaking the explanation given in this question is not correct: it is the non-leading (non-pivot) column on the LHS that shows the vectors are dependent, not the zero row. For example, if two vectors in ${\Bbb R}^3$ lead to an echelon form$$\pmatrix{1&2\cr0&3\cr0&0\cr}\ ,$$then the vectors are independent despite the zero row. If the LHS is a square matrix, as in the OP, it comes to the same thing, but usually it doesn't. $\endgroup$ – David Apr 23 '14 at 2:11
  • $\begingroup$ Corrected. I was admittedly dealing specifically with the case $n$ vectors in $\mathbb{R}^n$ (as in checking whether $n$ vectors form a basis of $\mathbb{R}^n$), but as this usually is not the case, I should not have left that as the main reason for seeing this. It just happens to work if the LHS is square. $\endgroup$ – Nicholas Stull Apr 23 '14 at 2:38
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Because

$$ 2(1,3,0) - (4,1,2) = (-2,5,-2) $$

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  • $\begingroup$ Was the method I used incorrect? How would I have gotten your answer? $\endgroup$ – hax0r_n_code Apr 23 '14 at 1:45
  • $\begingroup$ A basis is a set of vectors that span the vector space $V$ and is a linearly independent set. In this case $V = \mathbb{R}^3$, and your set of vectors is not linearly independent because of what I wrote. $\endgroup$ – user139708 Apr 23 '14 at 1:47
  • $\begingroup$ Yes, I understand what you're saying, but how would I have gotten to your answer since I didn't see it? Are there mechanics to finding this answer that you used or is it just something you spotted by examining the vectors? $\endgroup$ – hax0r_n_code Apr 23 '14 at 1:48
  • $\begingroup$ there is not a machine to do that. Just see your vectors and play with them. $\endgroup$ – user139708 Apr 23 '14 at 1:49
  • $\begingroup$ your solution is correct. You can put them in a matrix, and if the det is $0$, then they are not linearly independent $\endgroup$ – user139708 Apr 23 '14 at 1:51
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This is an amplification of Lemur's answer, expanding on the "see your vectors and play with them" in his comment. We want to know if we can find scalars $\alpha,\beta,\gamma$ such that $$\alpha(1,3,0)+\beta(4,1,2)+\gamma(-2,5,-2)=(0,0,0)\ .$$ But looking at the third component gives immediately $\beta=\gamma$. So the previous equation can be written as $$\alpha(1,3,0)+\beta(2,6,0)=(0,0,0)\ ,$$ and now it is easy to see that $\alpha=2$, $\beta=-1$ is a solution. So $$2(1,3,0)-(4,1,2)-(-2,5,-2)=(0,0,0)\ ,$$ which shows that the vectors are linearly dependent.

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