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?


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 $x_1\neq x_2\neq x_3$)

  • 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

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


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