# Thread: Linear Independent and Dependent

1. ## Linear Independent and Dependent

I need help understanding why the following is not linearly dependent
a=$\displaystyle \left[ \begin{array}{cccr}4\\1\\3\\\end{array} \right]$

b=$\displaystyle \left[ \begin{array}{cccr}2\\-1\\3\\\end{array} \right]$

c=$\displaystyle \left[ \begin{array}{cccr}-4\\2\\6\\\end{array} \right]$

Can someone explain the concepts on how to determined if it is linear independent or linear dependent?

P.S

2. Originally Posted by Paymemoney
I need help understanding why the following is not linearly dependent
a=$\displaystyle \left[ \begin{array}{cccr}4\\1\\3\\\end{array} \right]$

b=$\displaystyle \left[ \begin{array}{cccr}2\\-1\\3\\\end{array} \right]$

c=$\displaystyle \left[ \begin{array}{cccr}-4\\2\\6\\\end{array} \right]$

Can someone explain the concepts on how to determined if it is linear independent or linear dependent?

P.S
$\displaystyle \underline{\mbox{First way}}$ : Form the matrix $\displaystyle \begin {pmatrix}4&1&3\\2&\!\!\!-1&3\\\!\!\!-4&2&6\end {pmatrix}$ , which rows are the given vectors, and now calculate its determinant $\displaystyle \Delta$ . Since $\displaystyle \Delta\neq 0$ the matrix is regular and this happens iff its rows, seen as vectors, are linearly independent.

$\displaystyle \underline{\mbox{Second way}}$ : Suppose $\displaystyle a_1a+a_2b+a_3c=0\,,\,\,a_i\in\mathbb{F}=$ the field we're working with. You have to find out whether the coefficients HAVE TO BE zero or else the above equation is possible with at least one coefficient different from zero:

$\displaystyle 0=a_1a+a_2b+a_3c=a\begin {pmatrix}4\\1\\3\end {pmatrix}+b\begin {pmatrix}2\\\!\!\!-1\\3\end {pmatrix}+c\begin {pmatrix}\!\!-4\\\;2\\\;6\end {pmatrix}$ $\displaystyle =\begin {pmatrix}4a+2b-4c\\a-\;\;b+2c\\3a+3b+6c\end {pmatrix}$.
Now bring this matrix's coefficients into echelon form (reduce or prune the matrix using Gauss or Gauss-Jordan method) and you'll get that no row becomes all zeroes, which means the only possible solution to the above hmogeneous linear sistem is the trivial solution, i.e. $\displaystyle a_1=a_2=a_3=0$ and thus the vectors are lin. ind.

Tonio

3. The second way Tonio shows is the way I would prefer as it uses the definiton of "dependent" and "independent".

A set of vectors $\displaystyle \{v_1, v_2, \cdot\cdot\cdot, v_n\}$ is dependent if and only if there exist a set of scalars, $\displaystyle \{a_1, a_2, \cdot\cdot\cdot, a_n\}$, not all 0, such that $\displaystyle a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$.

A set of vectors is independent if it is not dependent- that is if the only way to have $\displaystyle a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$ is if $\displaystyle a_1= a_2= \cdot\cdot\cdot= a_n= 0$.

It is very important in mathematics to learn and use the exact words of definitions!