# Become a Linear Algebra Expert

For 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_1vec{v_1} + t_2vec{v_2} + … +t_nvec{v_n}

end{equation}

subsubsection{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.

subsubsection{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 (ref{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.

subsubsection{Examples}

$S = {begin{bmatrix}22end{bmatrix}, begin{bmatrix}1end{bmatrix}, begin{bmatrix}02end{bmatrix} $} is a set of Linearly Dependent vector.

The solution to the expression

begin{equation}

vec{0} = t_1begin{bmatrix}22end{bmatrix}+ t_2begin{bmatrix}1end{bmatrix} +t_3begin{bmatrix}02end{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 $T = {begin{bmatrix}1end{bmatrix}, begin{bmatrix}02end{bmatrix}$}.

The solution to the expression

begin{equation}

vec{0} = t_1begin{bmatrix}1end{bmatrix}+ t_2begin{bmatrix}01end{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.