0
$\begingroup$

If there is a group of vectors $v$ such that

$v=\left(\begin{array}{c} 1\\1 \end{array}\right), \left(\begin{array}{c} x_1\\x_2 \end{array}\right), \left(\begin{array}{c} x_1^2\\x_2^2 \end{array}\right)$

where $x_1\neq x_2,$ is $v$ linearly independent? If so, why?

Edit

Thank you! So what if

$v=\left(\begin{array}{c} 1\\1\\1 \end{array}\right), \left(\begin{array}{c} x_1\\x_2\\x_3 \end{array}\right), \left(\begin{array}{c} x_1^2\\x_2^2\\x_3^2 \end{array}\right), \left(\begin{array}{c} x_1^3\\x_2^3\\x_3^3 \end{array}\right)?$

(where $x_1\neq x_2\neq x_3$)

$\endgroup$
  • 1
    $\begingroup$ Not if $ x_1 = 1$ and $ x_2 = -1 $ $\endgroup$ – Ishfaaq Sep 16 '14 at 0:42
  • $\begingroup$ $3$ vectors with $2$ entries can never be linearly independent. Same goes for $4$ vectors with $3$ entries. $\endgroup$ – Omnomnomnom Sep 16 '14 at 1:34
  • $\begingroup$ @Omnomnomnom Why? $\endgroup$ – Dia McThrees Sep 16 '14 at 1:44
  • $\begingroup$ Rigorously speaking, it's a consequence of the dimension theorem $\endgroup$ – Omnomnomnom Sep 16 '14 at 1:46
  • $\begingroup$ @Omnomnomnom I'm sorry. I don't understand :( Could you explain? $\endgroup$ – Dia McThrees Sep 16 '14 at 2:06
1
$\begingroup$

No it is depend on x1 and x2 for example x1=0,x2=1 so v is dependent.

$\endgroup$

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.