Given the set

• Step 1: Set up a homogeneous system of equations
• Step 2: Transform the coefficient matrix of the system to the reduced row echelon form
• Step 3: Interpret the reduced row echelon form
• Comments

The set S = {v₁, v₂, v₃, v₄, v₅, v₆} of vectors in R⁴ is linearly independent if the only solution of

(*)    c₁v₁ + c₂v₂ + c₃v₃ + c₄v₄ + c₅v₅ + c₆v₆ = 0

is c₁, c₂, c₃, c₄, c₅, c₆ = 0.

In this case, the set S forms a basis for span S.

Otherwise (i.e., if a solution with at least some nonzero values exists), S is linearly dependent.

If this is the case, a subset of S can be found that forms a basis for span S.

With our vectors v₁, v₂, v₃, v₄, v₅, v₆, (*) becomes:

Rearranging the left hand side yields

The matrix equation above is equivalent to the following homogeneous system of equations

We now transform the coefficient matrix of the homogeneous system above to the reduced row echelon form to determine whether the system has

• the trivial solution only (meaning that S is linearly independent), or
• the trivial solution as well as nontrivial ones (S is linearly dependent).

The reduced row echelon form of the coefficient matrix of the homogeneous system (**) is

which corresponds to the system

The leading entries have been highlighted in yellow.

Those columns in the matrix that do not contain leading entries, correspond to unknowns that will be arbitrary. The system has infinitely many solutions:

Since the variables c₄, c₅ are arbitrary, then each of the vectors v₄, v₅ can be expressed as a linear combination of vectors in the set T = {v₁, v₂, v₃, v₆}. For example, set c₄ = 1, c₅ = 0, and use the equation (*) to express v₄ as a linear combination of the remaining vectors in the set S:

v₄ = 0v₁ +1v₂ +1v₃ +0v₆

Since the set T = {v₁, v₂, v₃, v₆} is linearly independent and it spans span S, then the set

forms a basis for span S.

• Shortcut: The pattern of leading entries can usually be determined before the reduced row echelon form is obtained (for example, a matrix in row echelon form is sufficient for this purpose). This is especially important when solving a problem like this by hand.