3
$\begingroup$

The exercise is:

Suppose $v_1, \ldots , v_m$ is linearly independent in $V$ and $w \in V$. Prove that if $v_1+w, \ldots, v_m+w$ is linearly dependent, then $w \in \operatorname{span}(v_1, \ldots, v_m)$.

My proof:
Since $v_1+w, \ldots, v_m+w$ is linearly dependent, we have for $a_1, \ldots, a_m \in F$ and not all equal to zero, $a_1(v_1+w) + \ldots + a_m(v_m+w) = 0$. If we let $c = -(a_1+ \ldots + a_m)$, we then find that $a_1v_1 + \ldots + a_m v_m = cw$.

First we consider the case where $c \neq 0$. Solving the equation above for $w$, we get $w = \frac{a_1}{c}v_1 + \ldots + \frac{a_m}{c}v_m$, making it an element of $\operatorname{span}(v_1, \ldots, v_m)$. If $c=0$, then we have $a_1v_1 + \ldots + a_mv_m = 0$, contradicting our first assumption that the list $v_1, \ldots, v_m$ is linearly independent. Therefore $c\neq0$, and the proof is complete.

Is this proof correct? How could I improve it? Thanks for any help.

$\endgroup$
1
  • 1
    $\begingroup$ Yes, it is correct. $\endgroup$
    – darij grinberg
    Jan 26, 2015 at 20:39

1 Answer 1

Reset to default
2
$\begingroup$

Yes, this is exactly the kind of argument I would expect to see (and I graded for a course that used this textbook). Note that the converse doesn't hold (take $w=0$).

$\endgroup$
2
  • $\begingroup$ The converse looks good to me. What do you think is wrong with it? $\endgroup$
    – Rob Arthan
    Jan 26, 2015 at 21:20
  • 1
    $\begingroup$ $0\in\operatorname{span}(v_1,\ldots,v_m)$, but $v_1+0,\ldots,v_m+0$ is linearly independent. $\endgroup$
    – Math1000
    Jan 26, 2015 at 21:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .