# Linear Independence with distinct variables

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 $x_1\neq x_2\neq x_3$)

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