Linear Independence
Table of contents
Linear Combination
Let $V$ be a vector space and finite number of vectors $\boldsymbol{x}_i \in V$ where $i \in \{1, \dots, k\}$.
$\forall \boldsymbol{v} \in V$ where $\exists \lambda_i \in \mathbb{R}$ such that:
\[\boldsymbol{v} = \sum_{i=1}^k \lambda_i \boldsymbol{x}_i\]Then $\boldsymbol{v}$ is a linear combination of $\boldsymbol{x}_i$.
Linear Combination for Zero Vector
The trivial linear combination for $\sum_{i=1}^k \lambda_i \boldsymbol{x}_i = \mathbf{0}$ is when $\forall \lambda_i = 0$.
If $\exists \lambda_i \neq 0$, then the linear combination is non-trivial.
Linearly Independent
Let $V$ be a vector space and $\boldsymbol{x}_i \in V$ where $i \in \{1, \dots, k\}$.
If there exists a non-trivial linear combination of $\boldsymbol{x}_i$ that equals $\mathbf{0}$, then $\boldsymbol{x}_i$ is linearly dependent.
Otherwise, if only a trivial combination exists, $\boldsymbol{x}_i$ is linearly independent.
If the only way to zero out the combination was to zero each and every one out, it means they have no relation to each other (no redundancy in information).
Interesting observations:
- If $\exists \boldsymbol{x}_i = \mathbf{0}$, then set of $\boldsymbol{x}_i$ is linearly dependent.
- Because setting all the others to zero and giving $\boldsymbol{x}_i$ a non-zero value is a non-trivial combination that equals $\mathbf{0}$.
- If any $\boldsymbol{x}_i$ is formed by a linear combination of the others, then set of $\boldsymbol{x}_i$ is linearly dependent.
- Because you can subtract the linear combination from $\boldsymbol{x}_i$, and the result is a non-trivial combination that equals $\mathbf{0}$.
- Gaussian elimination (just to row echelon) can be used to determine linear independence of the column vectors.
- Pivot columns are linearly independent to the columns to the left of them.
- Set of column vectors are linearly independent if all columns are pivot columns.