Linear Dependence

A set of Vectors are linearly dependant if each of them can be expressed as Linear Combinations of each other. 1

Definition §

Let $V_{\mathbb{F}}$ be a Vector Space fixed with Field $\mathbb{F}$. Let the scalars ${\alpha }_1,{\alpha }_2\dots {\alpha }_k\in F$ and the set of vectors be $S=\{{\vec{v}}_1,{\vec{v}}_2,\dots {\vec{v}}_k\}|{\vec{v}}_{1\dots k}\in V$

Linear Dependence §

$S$ is linearly dependant if and only if:

$$\sum _{i=1}^k{\alpha }_i{\vec{v}}_i=\vec{0}\text{ where at least one }{\alpha }_i\ne 0_F$$

In other words, if the zero vector can be represented as a Linear Combination of a set of vectors, where at least one scalar is non-zero, then the vectors are linearly dependant

Containing the zero vector

A set of vectors containing $\vec{0}$ is always linearly dependant.

Linear Independence §

$S$ is linearly independent if it is not linearly dependent. Alternatively, $S$ is linearly independent if and only if:

$$\sum _{i=1}^k{\alpha }_i{\vec{v}}_i=\vec{0}|a_{1\dots k}=0_F$$

In other words, the only way for a set of linearly independent vectors to sum up to the zero vector is if all of them are multiplied by 0.

Checking for Linear Independence §

Let $V_{\mathbb{F}}$ be a vector space fixed over field $\mathbb{F}$. Let $S_k$ be a set of vectors, $\{{\vec{v}}_1,{\vec{v}}_2,\dots {\vec{v}}_k\},{\vec{v}}_{1\dots k}\in V^n$

We can create a matrix, $\mathbf{M}$ of size $n\times k$ containing column matrices ${\mathbf{v}}_i$:

$$\mathbf{M}=[{\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_k]$$

$$\mathbf{M}=\left[\begin{array}{cccc}{{\mathbf{v}}_1}_1& {{\mathbf{v}}_2}_1& \dots & {{\mathbf{v}}_k}_1\\ {{\mathbf{v}}_1}_2& {{\mathbf{v}}_2}_2& \dots & {{\mathbf{v}}_k}_2\\ \dots & \dots & \dots & \dots \\ {{\mathbf{v}}_1}_n& {{\mathbf{v}}_2}_n& \dots & {{\mathbf{v}}_k}_n\end{array}\right]$$

If we augment a column matrix $0$ of size $n\times 1$ representing $\vec{0}\in V^n$

$$0=\left[\begin{array}{c}{\vec{0}}_1\\ {\vec{0}}_2\\ \dots \\ {\vec{0}}_n\end{array}\right]$$

We get:

$$[\mathbf{M}|0]$$

This is the Matrix Representation Of A Linear System and can be solved via Gauss-Jordan Elimination to obtain a RREF.

$S_k$ is a set of linearly independent vectors if and only if the matrix has a unique solution:

• $\text{rank}(\mathbf{M})=k$

If $k=n$ (resulting in $\mathbf{M}$ being a square matrix) then:

• $\det (\mathbf{M})\ne 0$

Theorem: Linear Dependence §

A set of $k$ vectors in the vector space $V^n$ is linearly dependent if $k>n$

Since $\mathbf{M}$ is a $n\times k$ matrix, the Matrix Rank of $\mathbf{M}$ is given by the smaller dimension i.e. $\text{rank}(\mathbf{M})=\text{min}(n,k)$

If $k>n$ then $\text{rank}(\mathbf{M})=n<k$

Therefore the vectors must be linearly dependant.