Linear Algebra: Linear Independence

Let A = { v₁, v₂, …, v_r} be a collection of vectors from Rⁿ. If r > 2 and at least one of the vectors in A can be written as a linear combination of the others, then A is said to be linearly dependent. The motivation for this description is simple: At least one of the vectors depends (linearly) on the others. On the other hand, if no vector in A is said to be a linearly independent set. It is also quite common to say that “the vectors are linearly dependent (or independent)” rather than “the set containing these vectors is linearly dependent (or independent).”

Example 1: Are the vectors v₁ = (2, 5, 3), v₂ = (1, 1, 1), and v₃ = (4, −2, 0) linearly independent?

If none of these vectors can be expressed as a linear combination of the other two, then the vectors are independent; otherwise, they are dependent. If, for example, v₃ were a linear combination of v₁ and v₂, then there would exist scalars k₁ and k₂ such that k v + k₂ v_2b = v₃. This equation reads

which is equivalent to

However, this is an inconsistent system. For instance, subtracting the first equation from the third yields k₁ = −4, and substituting this value into either the first or third equation gives k₂ = 12. However, ( k₁, k₂) = (−4, 12) does not satisfy the second equation. The conclusion is that v₃ is not a linear combination of v₁ and v₂. A similar argument would show that v₁ is not a linear combination of v₂ and v₃ and that v₂ is nota linear combination of v₁ and v₃. Thus, these three vectors are indeed linearly independent.

An alternative—but entirely equivalent and often simpler—definition of linear independence reads as follows. A collection of vectors v₁, v₂, …, v_r from Rⁿ is linearly independent if the only scalars that satisfy

are k₁ = k₂ = ⃛ = k_r = 0. This is called the trivial linear combination. If, on the other hand, there exists a nontrivial linear combination that gives the zero vector, then the vectors are dependent.

Example 2: Use this second definition to show that the vectors from Example 1— v₁ = (2, 5, 3), v₂ = (1, 1, 1), and v₃ = (4, −2, 0)—are linearly independent.

These vectors are linearly independent if the only scalars that satisfy

are k₁ = k₂ = k₃ = 0. But (*) is equivalent to the homogeneous system

Row-reducing the coefficient matrix yields

This echelon form of the matrix makes it easy to see that k₃ = 0, from which follow k₂ = 0 and k₁ = 0. Thus, equation (**)—and therefore (*)—is satisfied only by k₁ = k₂ = k₃ = 0, which proves that the given vectors are linearly independent.

Example 3: Are the vectors v₁ = (4, 1, −2), v₂ = (−3, 0, 1), and v₃ (1, −2, 1) linearly independent?

The equation k₁ v₁ + k₂ v₂ + k₃ v₃ = 0 is equivalent to the homogeneous system

Row-reduction of the coefficient matrix produces a row of zeros:

Since the general solution will contain a free variable, the homogeneous system (*) has nontrivial solutions. This shows that there exists a nontrivial linear combination of the vectors v₁, v₂, and v₃ that give the zero vector: v₁, v₂, and v₃ are dependent.

Example 4: There is exactly one value of c such that the vectors

are linearly dependent. Find this value of c and determine a nontrivial linear combination of these vectors that equals the zero vector.

As before, consider the homogeneous system

and perform the following elementary row operations on the coefficient matrix:

In order to obtain nontrivial solutions, there must be at least one row of zeros in this echelon form of the matrix. If c is 0, this condition is satisfied. Since c = 0, the vector v₄ equals (1, 1, 1, 0). Now, to find a nontrivial linear combination of the vectors v₁, v₂, v₃, and v₄ that gives the zero vector, a particular nontrivial solution to the matrix equation

is needed. From the row operations performed above, this equation is equivalent to

The last row implies that k₄ can be taken as a free variable; let k₄ = t. The third row then says

The second row implies

and, finally, the first row gives

Thus, the general solution of the homogeneous system (**)—and (*)—is

for any t in R. Choosing t = 1 , for example, gives k₁, k₂, k₃, k₄ so

is a linear combination of the vectors v₁, v₂, v₃, and v₄ that equals the zero vector. To verify that

simply substitute and simplify:

Infinitely many other nontrivial linear combinations of v₁, v₂, v₃, and v₄ that equal the zero vector can be found by simply choosing any other nonzero value of t in (***) and substituting the resulting values of k₁, k₂, k₃, and k₄ in the expression k₁ v₁ + k₂ v₂ + k₃ v₃ + k₄ v₄.

If a collection of vectors from Rⁿ contains more than n vectors, the question of its linear independence is easily answered. If C = { v₁, v₂, …, v_m} is a collection of vectors from Rⁿ and m > n, then C must be linearly dependent. To see why this is so, note that the equation

is equivalent to the matrix equation

Since each vector v_j contains n components, this matrix equation describes a system with m unknowns and n equations. Any homogeneous system with more unknowns than equations has nontrivial solutions, a result which applies here since m > n. Because equation (*) has nontrivial solutons, the vectors in C cannot be independent.

Example 5: The collection of vectors {2 i − j, i + j, − i + 4j} from R² is linearly dependent because any collection of 3 (or more) vectors from R² must be dependent. Similarly, the collection { i + j − k, 2i − 3j + k, i − 4k, − 2j,− 5i + j − 3k} of vectors from R³ cannot be independent, because any collection of 4 or more vectors from R³ is dependent.

Example 6: Any collection of vectors from Rⁿ that contains the zero vector is automatically dependent, for if { v₁, v₂,…, v_r−1, 0} is such a collection, then for any k ≠ 0,