Linear Independence

Think of a collection of vectors as a toolkit for building other vectors through linear combinations — expressions of the form $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k$, where the $c_i$ are real-number scalars.

A set is linearly independent when every tool pulls its own weight: no vector in the set can be written as a combination of the others. A set is linearly dependent when at least one vector is redundant — it duplicates directions already covered by the rest.

• In $\mathbb{R}^2$, the vectors $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ point in genuinely different directions — independent.
• The vectors $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 4 \end{bmatrix}$ lie on the same line (the second is twice the first) — dependent.

Here $\mathbb{R}^n$ denotes the space of column vectors with $n$ real entries, and $\mathbf{0}$ is the zero vector (all entries $0$).

The redundancy idea is made precise through a single equation.

Given vectors $\mathbf{v}_1,\dots,\mathbf{v}_k$ in $\mathbb{R}^n$, consider $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0}.$ Setting every coefficient $c_i = 0$ always works; this is the trivial solution.

• The set is linearly independent if the trivial solution is the only solution.
• The set is linearly dependent if there exists a nontrivial solution — scalars $c_1,\dots,c_k$, not all zero, satisfying the equation. Such an equation is called a dependency relation.

Why this captures redundancy: if, say, $c_j \neq 0$ in a dependency relation, you can divide by $c_j$ and solve for $\mathbf{v}_j$ as a combination of the others — so $\mathbf{v}_j$ really was redundant.

Two quick consequences of the definition:

• A set containing the zero vector is always dependent: put coefficient $1$ on $\mathbf{0}$ and $0$ on everything else to get a nontrivial relation.
• A single nonzero vector $\{\mathbf{v}\}$ is independent, since $c\mathbf{v}=\mathbf{0}$ forces $c=0$.

To test a concrete set, stack the vectors as the columns of a matrix $A = [\,\mathbf{v}_1\ \mathbf{v}_2\ \cdots\ \mathbf{v}_k\,]$. Then $c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k = A\mathbf{c}, \qquad \mathbf{c}=\begin{bmatrix} c_1 \\ \vdots \\ c_k \end{bmatrix},$ so the dependency equation becomes the homogeneous system $A\mathbf{c}=\mathbf{0}$.

Method 1 — row reduction. Row-reduce $A$. The columns are independent exactly when every column has a pivot (leading entry), i.e. there are no free variables. A free variable gives a nontrivial solution and hence dependence.

Method 2 — rank. The rank of $A$ is the number of pivots (equivalently, the number of linearly independent columns). The columns are independent iff $\operatorname{rank}(A) = k \ (\text{the number of vectors}).$ If $\operatorname{rank}(A) < k$, the set is dependent, and the number of free variables is $k - \operatorname{rank}(A)$.

Method 3 — determinant (square case only). If you have exactly $n$ vectors in $\mathbb{R}^n$, then $A$ is $n\times n$ and $\det(A) \neq 0 \iff \text{the columns are linearly independent}.$ A zero determinant signals dependence. This shortcut applies only when the number of vectors equals the dimension.

There is a hard ceiling on how many vectors can be independent at once.

Any set of more than $n$ vectors in $\mathbb{R}^n$ is automatically linearly dependent.

Reason: stacking $k > n$ vectors as columns gives an $n \times k$ matrix $A$. It has at most $n$ pivots (one per row), so at least $k - n \geq 1$ columns are free — there is always a nontrivial solution to $A\mathbf{c}=\mathbf{0}$.

Consequences you can use instantly:

• $3$ vectors in $\mathbb{R}^2$: dependent, no computation needed.
• $4$ vectors in $\mathbb{R}^3$: dependent.
• The maximum number of linearly independent vectors in $\mathbb{R}^n$ is exactly $n$.

Note the one-way nature: fewer than or equal to $n$ vectors may or may not be independent — you still have to test those.