Example 8: The trivial subspace, { 0}, of R ⁿ is said to have dimension 0. To be consistent with the definition of dimension, then, a basis for { 0} must be a collection containing zero elements; this is the empty set, ø.

The subspaces of R¹, R², and R³, some of which have been illustrated in the preceding examples, can be summarized as follows:

Example 9: Find the dimension of the subspace V of R⁴ spanned by the vectors

The collection { v₁, v₂, v₃, v₄} is not a basis for V—and dim V is not 4—because { v₁, v₂, v₃, v₄} is not linearly independent; see the calculation preceding the example above. Discarding v₃ and v₄ from this collection does not diminish the span of { v₁, v₂, v₃, v₄}, but the resulting collection, { v₁, v₂}, is linearly independent. Thus, { v₁, v₂} is a basis for V, so dim V = 2.

Example 10: Find the dimension of the span of the vectors

Since these vectors are in R⁵, their span, S, is a subspace of R⁵. It is not, however, a 3-dimensional subspace of R⁵, since the three vectors, w₁, w₂, and w₃ are not linearly independent. In fact, since w₃ = 3w₁ + 2w₂, the vector w₃ can be discarded from the collection without diminishing the span. Since the vectors w₁ and w₂ are independent—neither is a scalar multiple of the other—the collection { w₁, w₂} serves as a basis for S, so its dimension is 2.

The most important attribute of a basis is the ability to write every vector in the space in a unique way in terms of the basis vectors. To see why this is so, let B = { v₁, v₂, …, v _r } be a basis for a vector space V. Since a basis must span V, every vector v in V can be written in at least one way as a linear combination of the vectors in B. That is, there exist scalars k₁, k₂, …, k_r such that

To show that no other choice of scalar multiples could give v, assume that

Subtracting (*) from (**) yields

This expression is a linear combination of the basis vectors that gives the zero vector. Since the basis vectors must be linearly independent, each of the scalars in (***) must be zero:

Therefore, k′₁ = k₁, k′₂ = k₂,…, and k′_r = k_r, so the representation in (*) is indeed unique. When v is written as the linear combination (*) of the basis vectors v₁, v₂, …, v _r , the uniquely determined scalar coefficients k₁, k₂, …, k_r are called the components of v relative to the basis B. The row vector ( k₁, k₂, …, k_r ) is called the component vector of v relative to B and is denoted ( v) _B . Sometimes, it is convenient to write the component vector as a column vector; in this case, the component vector ( k₁, k₂, …, k_r )^T is denoted [ v] _B .

Example 11: Consider the collection C = { i, i + j, 2 j} of vectors in R². Note that the vector v = 3 i + 4 j can be written as a linear combination of the vectors in C as follows:

The fact that there is more than one way to express the vector v in R² as a linear combination of the vectors in C provides another indication that C cannot be a basis for R². If C were a basis, the vector v could be written as a linear combination of the vectors in C in one and only one way.

Example 12: Consider the basis B = { i + j, 2 i − j} of R². Determine the components of the vector v = 2 i − 7 j relative to B.

The components of v relative to B are the scalar coefficients k₁ and k₂ which satisfy the equation

This equation is equivalent to the system

The solution to this system is k₁ = −4 and k₂ = 3, so

Example 13: Relative to the standard basis { i, j, k} = { ê₁, ê₂, ê₃} for R³, the component vector of any vector v in R³ is equal to v itself: ( v) _B = v. This same result holds for the standard basis { ê₁, ê₂,…, ê_n} for every R ⁿ .

Orthonormal bases. If B = { v₁, v₂, …, v _n } is a basis for a vector space V, then every vector v in V can be written as a linear combination of the basis vectors in one and only one way:

Finding the components of v relative to the basis B—the scalar coefficients k₁, k₂, …, k_n in the representation above—generally involves solving a system of equations. However, if the basis vectors are orthonormal, that is, mutually orthogonal unit vectors, then the calculation of the components is especially easy. Here's why. Assume that B = {vˆ₁,vˆ₂,…,vˆ_n} is an orthonormal basis. Starting with the equation above—with vˆ₁, vˆ₂,…, vˆ_n replacing v₁, v₂, …, v _n to emphasize that the basis vectors are now assumed to be unit vectors—take the dot product of both sides with vˆ¹:

By the linearity of the dot product, the left-hand side becomes

Now, by the orthogonality of the basis vectors, vˆ_i · vˆ₁ = 0 for i = 2 through n. Furthermore, because vˆ is a unit vector, vˆ₁ · vˆ₁ = ‖vˆ₁‖1² = 1² = 1. Therefore, the equation above simplifies to the statement

In general, if B = { vˆ₁, vˆ₂,…, vˆ_n} is an orthonormal basis for a vector space V, then the components, k_i , of any vector v relative to B are found from the simple formula

Example 14: Consider the vectors

A nonzero vector is normalized—made into a unit vector—by dividing it by its length. Therefore,

Since B = { vˆ₁, vˆ₂, vˆ₃} is an orthonormal basis for R³, the result stated above guarantees that the components of v relative to B are found by simply taking the following dot products:

Therefore, ( v) _B = (5/3, 11/(3√2),3/√2), which means that the unique representation of v as a linear combination of the basis vectors reads v = 5/3 vˆ₁ + 11/(3√2) vˆ₂ + 3/√2 vˆ₃, as you may verify.

Example 15: Prove that a set of mutually orthogonal, nonzero vectors is linearly independent.

Proof. Let { v₁, v₂, …, v _r } be a set of nonzero vectors from some R ⁿ which are mutually orthogonal, which means that no v _i = 0 and v _i · v _j = 0 for i ≠ j. Let

The second equation follows from the first by the linearity of the dot product, the third equation follows from the second by the orthogonality of the vectors, and the final equation is a consequence of the fact that ‖ v₁‖² ≠ 0 (since v₁ ≠ 0). It is now easy to see that taking the dot product of both sides of (*) with v _i yields k_i = 0, establishing that every scalar coefficient in (*) must be zero, thus confirming that the vectors v₁, v₂, …, v _r are indeed independent.