Linear Independence and Dependence

Definition

Vectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$ are linearly independent if the only scalars making $c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_k \mathbf{v}_k = \mathbf{0}$

are $c_1 = c_2 = \cdots = c_k = 0$ (the trivial combination).

They are linearly dependent if there is some nontrivial combination (not all zero) giving $\mathbf{0}$ — equivalently, at least one of them is a linear combination of the others.

The matrix test

Form the matrix $M$ with the vectors as columns. Then:

• Independent ⟺ $M \mathbf{c} = \mathbf{0}$ has only the trivial solution ⟺ $M$ has rank equal to the number of vectors.
• Dependent ⟺ $M \mathbf{c} = \mathbf{0}$ has nontrivial solutions.

For a square $M$, this is the same as $\det M \ne 0$ (independent) vs. $\det M = 0$ (dependent).

Step-by-step — are the three vectors independent?

Set up $c_1 \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + c_2 \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} + c_3 \begin{pmatrix} 5 \\ 7 \\ 9 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

Matrix form $M \mathbf{c} = \mathbf{0}$: $M = \begin{pmatrix} 1 & 4 & 5 \\ 2 & 5 & 7 \\ 3 & 6 & 9 \end{pmatrix}$

Step 1 — Compute $\det M$ (expand along row 1): $\det M = 1 \cdot \det\begin{pmatrix} 5 & 7 \\ 6 & 9 \end{pmatrix} - 4 \cdot \det\begin{pmatrix} 2 & 7 \\ 3 & 9 \end{pmatrix} + 5 \cdot \det\begin{pmatrix} 2 & 5 \\ 3 & 6 \end{pmatrix}$

• $\det\begin{pmatrix} 5 & 7 \\ 6 & 9 \end{pmatrix} = 45 - 42 = 3$
• $\det\begin{pmatrix} 2 & 7 \\ 3 & 9 \end{pmatrix} = 18 - 21 = -3$
• $\det\begin{pmatrix} 2 & 5 \\ 3 & 6 \end{pmatrix} = 12 - 15 = -3$

$\det M = 1 \cdot 3 - 4 \cdot (-3) + 5 \cdot (-3) = 3 + 12 - 15 = 0$

Verdict: $\det M = 0$, so the vectors are linearly dependent.

Step 2 — Find the dependence relation. Observe that $\mathbf{v}_3 = (5, 7, 9) = (1, 2, 3) + (4, 5, 6) = \mathbf{v}_1 + \mathbf{v}_2$

Rearranged: $\mathbf{v}_1 + \mathbf{v}_2 - \mathbf{v}_3 = \mathbf{0}$, a nontrivial linear combination equal to zero. ✓

Geometric meaning

• In $\mathbb{R}^2$: two vectors are independent iff they are non-parallel.
• In $\mathbb{R}^3$: three vectors are independent iff they are not coplanar (don't all lie in the same plane through the origin).
• In $\mathbb{R}^n$: up to $n$ vectors can be independent; any $n + 1$ or more are always dependent.

If $\mathbf{v}_3 = \mathbf{v}_1 + \mathbf{v}_2$ (as here), the three vectors lie in the 2D plane $\operatorname{Span}\{\mathbf{v}_1, \mathbf{v}_2\}$.

Why independence matters

Independent vectors:

• Span uniquely — any vector in their span has exactly one linear combination expression.
• Form a basis for their span.
• Give a minimum-size spanning set — no redundancy.

Common mistakes

• Confusing "independence" with "orthogonality." Orthogonal implies independent, but independent vectors need not be orthogonal.
• Testing only one combination. Zero vector must be the only combination giving zero; one counterexample suffices for dependence.
• Counting vectors vs. dimensions. $n + 1$ vectors in $\mathbb{R}^n$ are automatically dependent, regardless of their specific values.

Try it in the visualization

Drag vector tips in 3D. When the three vectors become coplanar, the determinant shrinks to zero and the dependence indicator lights up. A sample dependence relation (e.g. $\mathbf{v}_3 = \mathbf{v}_1 + \mathbf{v}_2$) is printed.