Introduction

Linear independence helps us understand when a collection of vectors truly adds “new directions” to a space. When vectors are dependent, at least one of them can be built from the others — making it redundant.

This article assumes you already know the idea of span: the set of all linear combinations of a group of vectors.

What Does Linear Independence Mean?

A set of vectors is linearly independent if the only way to combine them to get the zero vector is by using all zero coefficients.

• For vectors $v_1, v_2, \dots, v_n$, consider the equation $$a_1 v_1 + a_2 v_2 + \dots + a_n v_n = 0.$$
• If the only solution is $$a_1 = a_2 = \dots = a_n = 0,$$ then the vectors are linearly independent.
• If there is any non‑zero solution, the vectors are dependent.

Interpretation:
A dependent vector can be written as a combination of the others — it adds no new direction.

Why Redundancy Happens

Redundancy appears when vectors “point the same way” or lie in the same span.

Common causes:

• Scalar multiples
  • Example: $(2,4)$ is just $2(1,2)$.
• One vector is a combination of others
  • Example: $(3,5)$ might equal $2(1,2) + 1(1,1)$.
• Too many vectors for the space
  • In $\mathbb{R}^2$, any set of 3 vectors is automatically dependent.

Redundant vectors don’t expand the span — they only repeat what the others already provide.

Geometric View

Thinking visually helps:

• In $\mathbb{R}^2$:
  • One non‑zero vector spans a line.
  • Two independent vectors span the whole plane.
  • A third vector must lie in that plane, so it cannot add anything new.
• In $\mathbb{R}^3$:
  • One vector → line
  • Two independent vectors → plane
  • Three independent vectors → full 3D space
  • A fourth vector → always dependent

Linear independence tells us how many “true directions” we have.

How to Check Linear Independence

Several simple methods work well at this level:

• Look for scalar multiples
  • If $v = k u$, they are dependent.
• Try to express one vector as a combination of others
  • Solve a small system of equations.
• Compare spans
  • If removing a vector doesn’t shrink the span, it was redundant.

Why Linear Independence Matters

Understanding independence helps with:

• Describing spaces efficiently
  • Use the smallest set of vectors that spans the space.
• Finding bases
  • A basis is a minimal spanning set — no redundancy.
• Solving systems of equations
  • Independent columns mean unique solutions.
• Understanding dimensions
  • The number of independent vectors equals the dimension of the span.

Examples

• $(1,2)$ and $(2,4)$
  • Dependent, because $(2,4) = 2(1,2)$.
• $(1,0)$ and $(0,1)$
  • Independent, and they span all of $\mathbb{R}^2$.
• $(1,1,0)$, $(0,1,1)$, $(1,0,1)$
  • These three are independent in $\mathbb{R}^3$.

Proving linear independence

Suppose we have three vectors in $\mathbb{R}^3$:

• $v_1 = (a_1, a_2, a_3)$
• $v_2 = (b_1, b_2, b_3)$
• $v_3 = (c_1, c_2, c_3)$

To check if they are linearly independent, we look at the equation $$x v_1 + y v_2 + z v_3 = (0,0,0)$$ and ask whether the only solution is $$x = 0,\quad y = 0,\quad z = 0.$$ If that’s the only way to get the zero vector, then $v_1, v_2, v_3$ are linearly independent.

Step 1: Write the system of equations

Write out the vector equation component‑wise: $$x(a_1, a_2, a_3) + y(b_1, b_2, b_3) + z(c_1, c_2, c_3) = (0,0,0).$$ This gives three equations: $$\begin{cases} x a_1 + y b_1 + z c_1 = 0 \\ x a_2 + y b_2 + z c_2 = 0 \\ x a_3 + y b_3 + z c_3 = 0 \end{cases}$$ So we have a system of 3 equations in the 3 unknowns $x, y, z$.

Step 2: View it as a matrix equation

We can write this as: $$\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$

• The columns of the matrix are the vectors $v_1, v_2, v_3$.
• We are checking whether the only solution is $x = y = z = 0$.

If yes, the vectors are linearly independent.

Step 3: A concrete example

Take:

• $v_1 = (1,0,1)$
• $v_2 = (0,1,1)$
• $v_3 = (1,1,2)$

We solve $$x(1,0,1) + y(0,1,1) + z(1,1,2) = (0,0,0).$$ Component‑wise: $$\begin{cases} x + 0y + z = 0 \\ 0x + y + z = 0 \\ x + y + 2z = 0 \end{cases}$$ So:

• First equation: $x + z = 0$
• Second equation: $y + z = 0$
• Third equation: $x + y + 2z = 0$

From the first two:

• $x = -z$
• $y = -z$

Substitute into the third: $$x + y + 2z = (-z) + (-z) + 2z = 0,$$ which is always true. So $z$ is free, and we get infinitely many solutions: $$(x,y,z) = (-z, -z, z).$$ Because there are non‑zero solutions (for example, $z = 1$ gives $(-1,-1,1)$), these vectors are linearly dependent.

Step 4: What linear independence would look like

For three vectors to be linearly independent, solving $$x v_1 + y v_2 + z v_3 = (0,0,0)$$ must force $$x = 0,\quad y = 0,\quad z = 0$$ and no other solutions.

In terms of the system:

• When you solve the 3 equations in $x, y, z$, you should get a unique solution.
• If you ever get a free variable (like $z$ above), then there are infinitely many solutions and the vectors are dependent.

So the method is:

• Set up $x v_1 + y v_2 + z v_3 = (0,0,0)$.
• Solve the resulting system for $x, y, z$.
• Conclude:
  • Only $(x,y,z) = (0,0,0)$ → vectors are linearly independent.
  • Any other solution exists → vectors are linearly dependent (one is redundant).

Calculator

Exercises

1. Determine whether the vectors $(3,6)$ and $(1,2)$ are linearly independent.

   Solution

   Dependent.
   $(3,6) = 3(1,2)$, so one is a scalar multiple of the other.
2. Are the vectors $(1,0,1)$, $(2,1,3)$, and $(1,1,2)$ linearly independent?

   Solution

   Independent.
   Solving $$a(1,0,1) + b(2,1,3) + c(1,1,2) = 0$$ gives only $a=b=c=0$.
3. Write $(4,5)$ as a linear combination of $(1,1)$ and $(1,2)$, if possible.

   Solution

   Yes.
   Solve $$a(1,1) + b(1,2) = (4,5).$$ This gives $a = 3$, $b = 1$.
4. Explain why any three vectors in $\mathbb{R}^2$ must be linearly dependent.

   Solution

   Because $\mathbb{R}^2$ has dimension 2.
   Any set with more vectors than the dimension must be dependent.
5. Determine whether $(1,2,3)$, $(2,4,6)$, and $(0,1,1)$ are linearly independent.

   Solution

   Dependent.
   $(2,4,6) = 2(1,2,3)$, so the first two are dependent.
6. Find coefficients $a$ and $b$ such that $$a(2,1) + b(1,3) = (5,7).$$

   Solution

   Solve $$2a + b = 5,\quad a + 3b = 7.$$ The solution is $a = 2$, $b = 1$.
7. True or false: If two vectors span $\mathbb{R}^2$, they must be linearly independent.

   Solution

   True.
   Spanning $\mathbb{R}^2$ requires two independent directions.
8. Determine whether the vectors $(1,1,1)$, $(2,3,4)$, and $(1,0,1)$ are linearly independent.

   Solution

   Independent.
   The only solution to $$a(1,1,1) + b(2,3,4) + c(1,0,1) = 0$$ is $a=b=c=0$.