# What are linearly dependent vectors like? [closed]

How are they different from linearly independent vectors?

• Hi and welcome to Math.SE. I suggest that you remove the integration tag, since it does not have to do with the question. You can do so by clicking the edit link below your question Also, what is your definition of linear independent and linear dependent set of vectors? Jun 17, 2016 at 16:14
• Can you tell us what kind of answer you want? Do you want a symbolic answer, with definitions? Do you want a more intuitive but perhaps less rigorous answer? (Oops, I guess the user account looks closed, perhaps?) Jun 17, 2016 at 16:32
• In simple terms: They are different from linearly independent vectors because for linearly dependent vectors you can find a linear combination of them that equates to zero, for linearly independent vectors you cannot. Jun 17, 2016 at 16:52

Linear dependent vectors $X_{1},X_{2},...,X_{n}$ arise when there exists a scalar $c_{1},c_{2},...,c_{n}$ such that :

$\sum_{i=1}^{n}c_{i}X_{i}=0$

If the determinant is 0, then the vectors are linearly dependent:

$$\begin{vmatrix} \ x_{11} & x_{12} & ...& x_{1n} \\ \ x_{21} & x_{22} & ... & x_{2n} \\ \ \vdots & \vdots & \ddots & \vdots \\ \ x_{n1} & x_{n2} & \cdots & x_{nn} \end{vmatrix} =0\notag$$

You can think of dependency as vectors having some relationship to each other (e.g. similar variables). If there is no scalar that exists then the vectors are linearly independent.

Example

$x_{1} = \begin{pmatrix} \ 1 \\ \ 1 \\ \ 1 \end{pmatrix}$ $x_{2} = \begin{pmatrix} \ 1 \\ \ -1 \\ \ 2 \\ \end{pmatrix}$ and $x_{3} = \begin{pmatrix} \ 3 \\ \ 1 \\ \ 4 \\ \end{pmatrix}$

These vectors are linearly dependent since $2x_{1} + x_{2} - x_{3} = 0$

Consider $\mathbb{R}^2$. Take the vectors $u= (1,1)$ and $v=(-2,-2)$. These vectors are called linearly dependent because they are equal up to a linear transformation, i.e. $v = (-2) \cdot u$.
This generalizes to $\mathbb{R}^n$ and more generally any kind of vector space using the definition given by Amstell above.