Let *A* = { **v** _{1}, **v** _{2}, …, **v** _{r }} be a collection of vectors from **R**^{n }. 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 **v** _{1} = (2, 5, 3), **v** _{2} = (1, 1, 1), and **v** _{3} = (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, **v** _{3} were a linear combination of **v** _{1} and **v** _{2}, then there would exist scalars *k* _{1} and *k* _{2} such that **k** **v** + **k** _{2} **v** _{2b} = **v** _{3}. This equation reads

which is equivalent to

However, this is an inconsistent system. For instance, subtracting the first equation from the third yields *k* _{1} = −4, and substituting this value into either the first or third equation gives *k* _{2} = 12. However, (*k* _{1}, *k* _{2}) = (−4, 12) does not satisfy the second equation. The conclusion is that **v** _{3} is not a linear combination of **v** _{1} and **v** _{2}. A similar argument would show that **v** _{1} is not a linear combination of **v** _{2} and **v** _{3} and that **v** _{2} is nota linear combination of **v** _{1} and **v** _{3}. 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 **v** _{1}, **v** _{2}, …, **v** _{r }from **R** ^{n }is linearly independent if the only scalars that satisfy are *k* _{1} = *k* _{2} = ⃛ = *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— **v** _{1} = (2, 5, 3), **v** _{2} = (1, 1, 1), and **v** _{3} = (4, −2, 0)—are linearly independent.

These vectors are linearly independent if the only scalars that satisfy

are *k* _{1} = *k* _{2} = *k* _{3} = 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 *k* _{3} = 0, from which follow *k* _{2} = 0 and *k* _{1} = 0. Thus, equation (**)—and therefore (*)—is satisfied only by *k* _{1} = *k* _{2} = *k* _{3} = 0, which proves that the given vectors are linearly independent.

**Example 3**: Are the vectors **v** _{1} = (4, 1, −2), **v** _{2} = (−3, 0, 1), and **v** _{3} (1, −2, 1) linearly independent?

The equation **k** _{1} **v** _{1} + **k** _{2} **v** _{2} + **k** _{3} **v** _{3} = **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 **v** _{1}, **v** _{2}, and **v** _{3} that give the zero vector: **v** _{1}, **v** _{2}, and **v** _{3} 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 **v** _{4} equals (1, 1, 1, 0). Now, to find a nontrivial linear combination of the vectors **v** _{1}, **v** _{2}, **v** _{3}, and **v** _{4} 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 *k* _{4} can be taken as a free variable; let *k* _{4} = *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 **k** _{1}, **k** _{2}, **k** _{3}, **k** _{4} so

is a linear combination of the vectors **v** _{1}, **v** _{2}, **v** _{3}, and **v** _{4} that equals the zero vector. To verify that

simply substitute and simplify:

Infinitely many other nontrivial linear combinations of **v** _{1}, **v** _{2}, **v** _{3}, and **v** _{4} that equal the zero vector can be found by simply choosing any other *nonzero* value of *t* in (***) and substituting the resulting values of *k* _{1}, *k* _{2}, *k* _{3}, and *k* _{4} in the expression **k** _{1} **v** _{1} + **k** _{2} **v** _{2} + **k** _{3} **v** _{3} + **k** _{4} **v** _{4}.

If a collection of vectors from **R** ^{n }contains more than *n* vectors, the question of its linear independence is easily answered. If *C* = { **v** _{1}, **v** _{2}, …, **v** _{m }} is a collection of vectors from **R** ^{n }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 **R** ^{2} is linearly dependent because *any* collection of 3 (or more) vectors from **R** ^{2} must be dependent. Similarly, the collection { **i** + **j** − **k**, **2i** − **3j** + **k**, **i** − **4k**, − **2j**,− **5i** + **j** − **3k**} of vectors from **R** ^{3} cannot be independent, because *any* collection of 4 or more vectors from **R** ^{3} is dependent.

**Example 6**: Any collection of vectors from **R**^{n }that contains the zero vector is automatically dependent, for if { **v** _{1}, **v** _{2},…, **v** _{r−1}, **0**} is such a collection, then for any *k* ≠ 0,

is a nontrivial linear combination that gives the zero vector.