# Become a Linear Algebra Expert

backFor a set of vector, $S = \{\vec{v_1},\vec{v_2},…,\vec{v_n}\}$, linear dependency is determined by the behaviour of this equation:

\begin{equation}

\label{eq:Lin}

\vec{0} = t_1\vec{v_1} + t_2\vec{v_2} + … +t_n\vec{v_n}

\end{equation}

0.0.1. Linearly Dependent
Set of vector, $S = \{\vec{v_1},\vec{v_2},…,\vec{v_n}\}$ is Linearly Dependent if there exists a solution to equation 1 where $t_1,…,t_n$ are \textbf{not all zeros}.

This also translates to this expression:

If a vector in the set can be expressed as a linear combination as other vector, then that set of vector is linearly dependent.

0.0.2. Linearly Independent
Opposite of linearly dependent, Set of vector, $S = \{\vec{v_1},\vec{v_2},…,\vec{v_n}\}$ is Linearly Independent if the only solution to (eq:Lin) is if $t_1,…,t_n$ are all zeros.

This also translates to this expression:

If any vector in the set is not possible to be expressed as a linear combination of other vectors in the set, then the set of vectors is linearly independent.

0.0.3. Examples
$S = \{\begin{bmatrix}2\\2\end{bmatrix}, \begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\2\end{bmatrix} $\} is a set of Linearly Dependent vector.

The solution to the expression

\begin{equation}

\vec{0} = t_1\begin{bmatrix}2\\2\end{bmatrix}+ t_2\begin{bmatrix}1\\0\end{bmatrix} +t_3\begin{bmatrix}0\\2\end{bmatrix}

\end{equation}

can be solved by $t_1 = 1, t_2=-2, t_3=-1$. Since $t_i$ values are not all zeros, this set of vector is linearly dependent.

Consider $T = \{\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\2\end{bmatrix}$\}.

The solution to the expression

\begin{equation}

\vec{0} = t_1\begin{bmatrix}1\\0\end{bmatrix}+ t_2\begin{bmatrix}0\\1\end{bmatrix}

\end{equation}

is only the trivial solution, where all the $t_i$ values must be zero in order for the equation to be satisfied. Therefore, set of vector $T$ is linearly independent.