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Linear
Independence

Discover when vectors carry unique directional information and when they are merely redundant copies of each other.

Linear Algebra

What Is a Linear Combination?

A linear combination of vectors \(\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n\) is any expression of the form:

Think of scalars as volume knobs: they stretch, shrink, or flip each vector before you add them together. The result is a new vector built from the originals.

The Big Question

Given a set of vectors, we want to know: can any one of them be written as a linear combination of the others?

Independent
No vector is redundant: each adds a new direction to the set.
Dependent
At least one vector is a copy or blend of the others: it's redundant.

Visual Intuition in \(\mathbb{R}^2\)

Independent \(\mathbf{v}_1\) \(\mathbf{v}_2\) Span = all of \(\mathbb{R}^2\) Dependent \(\mathbf{v}_1\) \(\mathbf{v}_2\) Span = just a line

The Formal Definition

The equation \(c_1\mathbf{v}_1 + \cdots + c_n\mathbf{v}_n = \mathbf{0}\) is called a linear dependence relation. Everything comes down to whether the zero vector can be built in a non-trivial way.

The Matrix Connection

Place the vectors as columns of a matrix \(A\). Then the dependence equation becomes the homogeneous system:

Independent
\(A\mathbf{x}=\mathbf{0}\) has only \(\mathbf{x}=\mathbf{0}\): every column is a pivot column.
Dependent
\(A\mathbf{x}=\mathbf{0}\) has free variables: at least one column is not a pivot column.

Row-reduce \(A\) to echelon form: if every column has a pivot, the vectors are independent.

Independence vs Dependence

Linearly Independent Linearly Dependent
Equation \(c_1\mathbf{v}_1+\cdots=\mathbf{0}\) Only trivial solution Non-trivial solutions exist
Homogeneous system No free variables At least one free variable
Geometric meaning Vectors span new directions A vector lies in others' span
Redundancy No vector is removable At least one is removable
Determinant (square) \(\det(A) \neq 0\) \(\det(A) = 0\)

How to Test for Independence

1
Form the matrix. Place the vectors as columns of a matrix \(A\).
2
Row-reduce. Apply Gaussian elimination to reach echelon form.
3
Count pivots. If every column contains a pivot: independent. If any column lacks a pivot: dependent.
4
Shortcut for square matrices: compute \(\det(A)\). Non-zero means independent.

Example 1: Independent in \(\mathbb{R}^2\)

Are \(\mathbf{v}_1 = \begin{pmatrix}1\\2\end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix}3\\1\end{pmatrix}\) linearly independent?

Example 2: Dependent in \(\mathbb{R}^2\)

Are \(\mathbf{v}_1 = \begin{pmatrix}2\\4\end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix}1\\2\end{pmatrix}\) linearly independent?

Example 3: Three Vectors in \(\mathbb{R}^3\)

Test independence of \(\mathbf{v}_1=\begin{pmatrix}1\\0\\2\end{pmatrix}\), \(\mathbf{v}_2=\begin{pmatrix}0\\1\\1\end{pmatrix}\), \(\mathbf{v}_3=\begin{pmatrix}2\\1\\5\end{pmatrix}\).

Visualizing in \(\mathbb{R}^3\)

x y z \(\mathbf{v}_1\) \(\mathbf{v}_2\) \(\mathbf{v}_3\) \(\mathbf{v}_3 = 2\mathbf{v}_1 + \mathbf{v}_2\) All three vectors lie in the same plane

Key Theorems

1
More vectors than dimensions: if you have more than \(n\) vectors in \(\mathbb{R}^n\), they must be linearly dependent.
2
Zero vector rule: any set containing the zero vector \(\mathbf{0}\) is automatically dependent.
3
Two-vector test: two vectors are dependent if and only if one is a scalar multiple of the other.
4
Determinant test (square): \(n\) vectors in \(\mathbb{R}^n\) are independent iff \(\det([\mathbf{v}_1 \cdots \mathbf{v}_n]) \neq 0\).

Why It Matters

Bases
A basis is a linearly independent set that spans a space: the minimal "coordinate system."
Unique Solutions
\(A\mathbf{x}=\mathbf{b}\) has a unique solution iff the columns of \(A\) are independent.
Dimension
The number of vectors in any basis equals the dimension of the subspace.

Common Mistakes

✕ Confusing the Zero Vector
Students forget that \(\{\mathbf{v}, \mathbf{0}\}\) is always dependent because \(1\cdot\mathbf{0}+0\cdot\mathbf{v}=\mathbf{0}\).
✕ Checking Pairs Only
Three vectors can be dependent even when no two are multiples of each other. You must test the full set together.
✕ Rows vs Columns
When forming the matrix, vectors go as columns, not rows. Putting them as rows tests a different question.

Question 1 of 5

Are the vectors \(\mathbf{v}_1 = \begin{pmatrix}1\\3\end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix}2\\6\end{pmatrix}\) linearly independent or dependent?

Question 2 of 5

Which of the following sets is always linearly dependent?

Question 3 of 5

You row-reduce the matrix \(A = [\mathbf{v}_1\ \mathbf{v}_2\ \mathbf{v}_3]\) and obtain:

What can you conclude?

Question 4 of 5

The set \(\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{0}\}\) is:

Question 5 of 5

If \(\det\begin{pmatrix}3&1\\6&2\end{pmatrix} = 0\), what does this tell us about the columns?