Let $u = (1, 2, 0, 1)$, $v = (1, −1, 1, 0)$, $w = (2, 4, 0, 2)$, and suppose $W = \text{span}(u, v, w)$ is a subspace of $\mathbb{R}^4$. Is $u,v,w$ a basis of $W$? Why or why not?

I'm not sure if it is or not, I reduced it down to echelon form. $$ \left[ \begin{array}{cccc} 1&2&0&1\\ 0&1&-1/3&1/3\\ 0&0&0&0 \end{array} \right] $$

But I'm not sure where to go from here.

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    $\begingroup$ The bottom row consists of all zeros. What does that tell you about whether $u, v, w$ are linearly independent? What would have to be true for $u, v, w$ to be linearly independent. $\endgroup$ – amWhy Mar 25 '17 at 20:53

Recall that a set $\mathcal{B}$ is a basis for $W$ if $Span\mathcal{B}=W$ and $\mathcal{B}$ is linearly independent. In your problem, $\{u,v,w\}$ spans $W$ by definition of $W$. So you only have to check the linear (in)dependence of $\{u,v,w\}$. How can you relate linear independence of rows of a matrix to the number of rows that have all zero entries in the matrix's reduced echelon form?

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    $\begingroup$ Since SpanB is not linear independent because of the zero row than it is not a basis for W right? $\endgroup$ – Blare Mar 25 '17 at 20:51
  • $\begingroup$ You got it, Blare! $\endgroup$ – amWhy Mar 25 '17 at 20:54
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    $\begingroup$ @Blare: you have the right idea! But be careful with your terms. $\mathcal B$ is a set of vectors, while $\operatorname{Span} \mathcal B$ is a vector space. A vector space can't be linearly (in)dependent. What you mean is that $\mathcal B$ is linearly dependent. $\endgroup$ – Matthew Leingang Mar 26 '17 at 0:06
  • $\begingroup$ @MatthewLeingang Question does it matter that W is a subspace of R4 and what if all three vectors were linear independent would they form a basis even though there are only 3 vectors? Thank you very much! $\endgroup$ – Blare Mar 26 '17 at 20:00
  • $\begingroup$ @Blare: The dimension of the “ambient” space ($\mathbb{R}^4$ in this case) doesn't matter, except that three vectors in $\mathbb{R}^2$ can't be linearly independent. The answer to your second question is yes: A basis of a vector (sub)space is a linearly independent spanning set. $W$ is, by construction, spanned by $u$, $v$, and $w$, so if those three are linearly independent, they form a basis for $W$. $\endgroup$ – Matthew Leingang Mar 27 '17 at 16:22


The three vectors are a basis iff they are linearly independent, but
a simple inspection shows that $w=2u$ (as confirmed by your echelon form reduction), so ......

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  • $\begingroup$ I only edited (added one letter) to change "tree" to "three" vectors... $\endgroup$ – amWhy Mar 25 '17 at 20:55
  • $\begingroup$ Well, your mastery of English is exceptional! I've got a sticky "e" key on my keyboard, which requires a forceful finger to render. So I've been guilty of typing "th" instead of "the" etc. No need to worry about your English!...Typos happen!! $\endgroup$ – amWhy Mar 25 '17 at 21:09

If $u, v, w $ are linearly independent and span $ W $, then they form a basis.

But you can see that they are not linearly independent. As your echelon form suggests, $w$ could be turned into a zero vector by the linear combination of $u$ and $v$.

For the definition of linear independence, check this wiki article.

More precisely,

$w$ = 2$u$

This means that $w$ is redundant and you only need two vectors i.e. $u$ and $v$ to describe every vector in the subspace $W$.

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