5
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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
9
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
4
$\begingroup$

Hint:

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 ......

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $W$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.