Vectors \(\overrightarrow{v}_1, \overrightarrow{v}_2, \dots, \overrightarrow{v}_n\) of the same dimension are linearly independent if \[a_1 \overrightarrow{v}_1 + a_2 \overrightarrow{v}_2 + \dots + a_n \overrightarrow{v}_n = \overrightarrow{0}\] implies all the scalars \(a_1, a_2, \dots, a_n\) are \(0.\)
The vectors \(\overrightarrow{v}_1 = < 1,2 >\) and \(\overrightarrow{v}_2 = < 3, 6 >\) lie on the same line so intuitively they are not linearly independent. Using the definition given above, we can show they are not linearly independent by finding \(a_1 \neq 0\) and \(a_2 \neq 0\) such that \[a_1 \overrightarrow{v}_1 + a_2 \overrightarrow{v}_2 = \overrightarrow{0}\] One solution is \(a_1 = -3\) and \(a_2 = 1.\) Plugging in those values, we have \[-3 < 1,2 > + < 3, 6 > = < 0, 0 >\] So, \(< 1,2 >\) and \(< 3, 6 >\) are linearly dependent.
The vectors \(\overrightarrow{v}_1 = < 1,0,0,0 >,\) \(\overrightarrow{v}_2 = < 1,2,0,0 >\) and \(\overrightarrow{v}_3 = < 1,2,0,3 >\) are linearly independent \(4\)-dimensional vectors. To check this, suppose \[a_1 \overrightarrow{v}_1 + a_2 \overrightarrow{v}_2 + a_3 \overrightarrow{v}_3 = \overrightarrow{0}\] The first coordinate implies \(a_1 + a_2 + a_3 = 0.\) The second coordinate implies \(2a_2 + 2a_3 = 0.\) The third coordinate implies \(0 = 0\) which gives no information. The fourth coordinate implies \(3a_3 = 0.\)
The equation yielded from the fourth coordinate implies \(a_3 = 0.\) Plugging \(a_3 = 0\) into the equation yielded by the second coordinate gives \(2a_2 = 0,\) so \(a_2 = 0.\) Plugging in \(a_2 = 0\) and \(a_3 = 0\) into the equation yielded by the first coordinate gives \(a_1 = 0.\) Thus, the only solution is \(a_1 = a_2 = a_3 = 0\) and we can conclude that \(\overrightarrow{v}_1, \overrightarrow{v}_2\) and \(\overrightarrow{v}_3\) are linearly independent.
The vectors \(\overrightarrow{v}_1 = < 1,1,2 >,\) \(\overrightarrow{v}_2 = < 2,0,-1 >\) and \(\overrightarrow{v}_3 = < 4,2,3 >\) are not linearly independent \(3\)-dimensional vectors.
In particular, \(2\overrightarrow{v}_1 + \overrightarrow{v}_2 = \overrightarrow{v}_3.\) By subtracting \(\overrightarrow{v}_3\) from both sides, we get \[2\overrightarrow{v}_1 + \overrightarrow{v}_2 - \overrightarrow{v}_3 = 0\] This is a solution to \[a_1\overrightarrow{v}_1 + a_2\overrightarrow{v}_2 + a_3\overrightarrow{v}_3 = 0\] where \(a_1, a_2,\) and \(a_3\) are not all \(0.\)