Linear Independence Calculator

Enter 2 or 3 vectors in ℝ² or ℝ³ to test linear independence using matrix rank (Gaussian elimination) and, for square systems, the determinant test.

Quick Facts

Independence definition
c₁v₁+...+cₖvₖ=0 ⇒ all cᵢ=0
Only the trivial combination of scalars should produce the zero vector.
Rank test (general)
independent iff rank = k
Row-reduce the vectors; the rank is the number of nonzero pivot rows.
Determinant test (square only)
independent iff det ≠ 0
Only applies when the number of vectors equals the dimension.
Maximum independent set
at most n vectors in ℝⁿ
More than n vectors in ℝⁿ are always linearly dependent.

Your Results

Calculated
Rank
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Number of independent directions found
Determinant
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Only defined when vectors = dimension
Independence
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Independent vs. dependent verdict
Interpretation
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What the result means

Ready

Choose the number of vectors and dimension, enter the components, then press Calculate.

How to Determine Linear Independence

A set of vectors v₁, v₂, ..., vₖ is linearly independent if the only way to combine them with scalars c₁, c₂, ..., cₖ to get the zero vector is the trivial combination: c₁v₁ + c₂v₂ + ... + cₖvₖ = 0 forces c₁ = c₂ = ... = cₖ = 0. If any other combination (not all zero) also produces the zero vector, the set is linearly dependent — at least one vector can be written as a linear combination of the others. This calculator checks 2 or 3 vectors in ℝ² or ℝ³ using the general rank test and, when the system is square, the determinant shortcut.

How the calculation works

The calculator stacks your vectors as the rows of a matrix and reduces that matrix to row-echelon form using Gaussian elimination with partial pivoting (swapping in the largest available pivot at each step for numerical stability). The rank is the number of nonzero pivot rows left after elimination. If the rank equals the number of vectors k, every vector contributes a new, independent direction and the set is linearly independent; if the rank is less than k, some vectors are redundant and the set is dependent. When the number of vectors equals the dimension (a square system), the calculator also reports the determinant: for 2 vectors in ℝ², det = a·d − b·c; for 3 vectors in ℝ³, the calculator expands along the first row using the standard 3×3 cofactor formula. A nonzero determinant is equivalent to a full rank and confirms independence; a zero determinant confirms dependence.

Common mistakes

  • Applying the determinant test to a non-square system: the determinant is only defined for a square matrix, so you can only use it directly when the number of vectors equals the dimension (e.g., 2 vectors in ℝ² or 3 vectors in ℝ³). For any other combination, use the rank test instead.
  • Assuming more vectors always add information: in an n-dimensional space, any set of more than n vectors is automatically dependent, since the rank can never exceed n. Three vectors in ℝ² will always be dependent.
  • Confusing "nonzero vectors" with "independent vectors": two nonzero vectors can still be dependent if one is a scalar multiple of the other (they point along the same line).

Real-world applications

  • Solving linear systems: a square coefficient matrix has a unique solution exactly when its rows (or columns) are linearly independent, i.e. its determinant is nonzero.
  • Basis and dimension: a set of n linearly independent vectors in ℝⁿ forms a basis — every other vector in the space can be written uniquely as a combination of them.
  • Computer graphics and physics: independent direction vectors (e.g., for coordinate axes or force components) avoid redundant or degenerate transformations.
  • Data science: checking whether feature vectors are independent helps detect multicollinearity before fitting a linear model.

Frequently Asked Questions

What does it mean for a set of vectors to be linearly independent?
A set of vectors v₁, ..., vₖ is linearly independent if the only solution to c₁v₁ + c₂v₂ + ... + cₖvₖ = 0 is c₁ = c₂ = ... = cₖ = 0. If some other combination of scalars (not all zero) also produces the zero vector, the set is linearly dependent — meaning at least one vector can be written as a combination of the others.
How do you test linear independence with a determinant?
For n vectors in ℝⁿ, arrange them as the rows (or columns) of an n×n matrix and compute its determinant. If the determinant is nonzero, the vectors are linearly independent and form a basis for ℝⁿ. If the determinant equals zero, they are linearly dependent. This shortcut only works when the number of vectors equals the dimension.
What is the rank test for linear independence, and why use it for non-square cases?
The rank test works for any number of vectors in any dimension: reduce the vectors (as rows of a matrix) to row-echelon form via Gaussian elimination and count the nonzero pivot rows — that count is the rank. If the rank equals the number of vectors, they're linearly independent; if the rank is smaller, they're dependent. Unlike the determinant test, the rank test also handles cases where the vector count differs from the dimension.
Can more vectors than the dimension ever be linearly independent?
No. In an n-dimensional space, any set of more than n vectors is automatically linearly dependent, because the rank of the matrix can never exceed n (the number of columns/dimension). For example, any 3 vectors in ℝ² are guaranteed to be linearly dependent.