The familiar coordinate plane ${\mathbf{\text{R}}}^2$ is the set of all ordered pairs (a, b) of real numbers. We may regard ${\mathbf{\text{R}}}^2$ as the xy-plane in ${\mathbf{\text{R}}}^3$ by identifying (a, b) with the vector (a, b, 0) in ${\mathbf{\text{R}}}^3$. Then ${\mathbf{\text{R}}}^2$ is simply the set of all 3-dimensional vectors that have third component 0.

Clearly, the sum of any two vectors in ${\mathbf{\text{R}}}^2$ is again a vector in ${\mathbf{\text{R}}}^2$, as is any scalar multiple of a vector in ${\mathbf{\text{R}}}^2$. Indeed, vectors in ${\mathbf{\text{R}}}^2$ satisfy all the properties of a vector space enumerated in Theorem 1. Consequently, the plane ${\mathbf{\text{R}}}^2$ is a vector space in its own right.

The two vectors u and v are collinear—they lie on the same line through the origin and hence point either in the same direction (Fig. 4.1.5) or in opposite directions—if and only if one is a scalar multiple of the other; that is, either

for some scalar c. The scalar c merely adjusts the length and direction of one vector to fit the other. If u and v are nonzero vectors, then $c=\pm |\mathbf{\text{u}}|/|\mathbf{\text{v}}|$ in the first relation, with c being positive if the two vectors point in the same direction, negative otherwise.

If one of the relations in (4) holds for some scalar c, then we say that the two vectors are linearly dependent. Note that if $\mathbf{\text{u}}=\mathbf{0}$ while $\mathbf{\text{v}}\ne \mathbf{0},$ then $\mathbf{\text{u}}=0\mathbf{\text{v}}$ but v is not a scalar multiple of u. (Why?) Thus, if precisely one of the two vectors u and v is the zero vector, then u and v are linearly dependent, but only one of the two relations in (4) holds.

If u and v are linearly dependent vectors with $\mathbf{\text{u}}=c\mathbf{\text{v}}$ (for instance), then $1\cdot \mathbf{\text{u}}+(-c)\cdot \mathbf{\text{v}}=\mathbf{0}.$ Thus there exist scalars a and b not both zero such that

Conversely, suppose that Eq. (5) holds with a and b not both zero. If $a\ne 0$ (for instance) then we can solve for

with $c=-b/a$, so it follows that u and v are linearly dependent. Therefore, we have proved the following theorem.

The most interesting pairs of vectors are those that are not linearly dependent. The two vectors u and v are said to be linearly independent provided that they are not linearly dependent. Thus u and v are linearly independent if and only if neither is a scalar multiple of the other. By Theorem 2 this is equivalent to the following statement:

Thus the vectors u and v are linearly independent provided that no nontrivial linear combination of them is equal to the zero vector.

The most important property of linearly independent pairs of plane vectors is this: If u and v are linearly independent vectors in the plane, then any third vector w in ${\mathbf{\text{R}}}^2$ can be expressed as a linear combination of u and v. That is, there exist scalars a and b such that $\mathbf{\text{w}}=a\mathbf{\text{u}}+b\mathbf{\text{v}}$ (Fig. 4.1.6). This is a statement that the two linearly independent vectors u and v suffice (in an obvious sense) to “generate” the whole plane ${\mathbf{\text{R}}}^2$. This general fact—which Section 4.3 discusses in a broader context—is illustrated computationally by the following example.

We have said that the two vectors are linearly dependent provided that they lie on the same line through the origin. For three vectors $\mathbf{\text{u}}=(u_1,u_2,u_3),\mathbf{\text{v}}=(v_1,v_2,v_3)$, and $\mathbf{\text{w}}=(w_1,w_2,w_3)$ in space, the analogous condition is that the three points $(u_1,u_2,u_3),(v_1,v_2,v_3)$, and $(w_1,w_2,w_3)$ lie in the same plane through the origin in ${\mathbf{\text{R}}}^3$. Given u, v, and w, how can we determine whether the vectors u, v, and w are coplanar? The key to the answer is the following observation: If r and s are scalars, then the parallelogram law of addition implies that the vectors u, v, and $r\mathbf{\text{u}}+s\mathbf{\text{v}}$ are coplanar; specifically, they lie in the plane through the origin that is determined by the parallelogram with vertices 0, ru, sv, and $r\mathbf{\text{u}}+s\mathbf{\text{v}}.$ Thus any linear combination of u and v is coplanar with u and v. This is the motivation for our next definition.

Note that each of the three equations in (6) implies that there exist three scalars a, b, and c not all zero such that

For if $\mathbf{\text{w}}=r\mathbf{\text{u}}+s\mathbf{\text{v}}$ (for instance), then

so we can take $a=r,b=s$, and $c=-1\ne 0$. Conversely, suppose that (7) holds with a, b, and c not all zero. If $c\ne 0$ (for instance), then we can solve for

with $r=-a/c$ and $s=-b/c$, so it follows that the three vectors u, v, and w are linearly dependent. Therefore, we have proved the following theorem.

The three vectors u, v, and w are called linearly independent provided that they are not linearly dependent. Thus u, v, and w are linearly independent if and only if neither of them is a linear combination of the other two. As a consequence of Theorem 3, this is equivalent to the following statement:

Thus the three vectors u, v, and w are linearly independent provided that no nontrivial linear combination of them is equal to the zero vector.

Given two vectors, we can see at a glance whether either is a scalar multiple of the other. By contrast, it is not evident at a glance whether or not three given vectors in ${\mathbf{\text{R}}}^3$ are linearly independent. The following theorem provides one way to resolve this question.

Hence, in order to determine whether or not three given vectors u, v, and w are linearly independent, we can calculate the determinant in (8). In practice, however, it is usually more efficient to set up and solve the linear system in (9). If we obtain only the trivial solution $a=b=c=0$, then the three given vectors are linearly independent. But if we find a nontrivial solution, we then can express one of the vectors as a linear combination of the other two and thus see how the three vectors are linearly dependent.

Perhaps the most familiar triple of linearly independent vectors in ${\mathbf{\text{R}}}^3$ consists of the basic unit vectors

When represented by arrows with the initial points at the origin, these three vectors point in the positive directions along the three coordinate axes (Fig. 4.1.7). The expression

shows both that

• the three vectors i, j, and k are linearly independent (because $\mathbf{\text{v}}=\mathbf{0}$ immediately implies $a=b=c=0$), and that

• any vector in ${\mathbf{\text{R}}}^3$ can be expressed as a linear combination of i, j, and k.

A basis for ${\mathbf{\text{R}}}^3$ is a triple u, v, w of vectors such that every vector t in ${\mathbf{\text{R}}}^3$ can be expressed as a linear combination

of them. That is, given any vector t in ${\mathbf{\text{R}}}^3$, there exist scalars a, b, c such that Eq. (11) holds. Thus the unit vectors i, j, and k constitute a basis for ${\mathbf{\text{R}}}^3$. The following theorem says that any three linearly independent vectors constitute a basis for ${\mathbf{\text{R}}}^3$.

Up until this point, we have used the words line and plane only in an informal or intuitive way. It is now time for us to say precisely what is meant by a line or plane through the origin in ${\mathbf{\text{R}}}^3$. Each is an example of a subspace of ${\mathbf{\text{R}}}^3$.

The nonempty subset V of ${\mathbf{\text{R}}}^3$ is called a subspace of ${\mathbf{\text{R}}}^3$ provided that V itself is a vector space under the operations of vector addition and multiplication of vectors by scalars. Suppose that the nonempty subset V of ${\mathbf{\text{R}}}^3$ is closed under these operations—that is, that the sum of any two vectors in V is also in V and that every scalar multiple of a vector in V is also in V. Then the vectors in V automatically satisfy properties (a) through (h) of Theorem 1, because these properties are “inherited” from ${\mathbf{\text{R}}}^3$; they hold for all vectors in ${\mathbf{\text{R}}}^3$, including those in V. Consequently, we see that a nonempty subset V of ${\mathbf{\text{R}}}^3$ is a subspace of ${\mathbf{\text{R}}}^3$ if and only if it satisfies the following two conditions:

1. If u and v are vectors in V, then $\mathbf{\text{u}}+\mathbf{\text{v}}$ is also in V (closure under addition).

2. If u is a vector in V and c is a scalar, then cu is in V (closure under multiplication by scalars).

It is immediate that $V={\mathbf{\text{R}}}^3$ is a subspace: ${\mathbf{\text{R}}}^3$ is a subspace of itself. At the opposite extreme, the subset $V=\{\mathbf{0}\}$, containing only the zero vector, is also a subspace of ${\mathbf{\text{R}}}^3$, because $\mathbf{0}+\mathbf{0}=\mathbf{0}$ and $c\mathbf{0}=\mathbf{0}$ for every scalar c. Thus $V=\{\mathbf{0}\}$ satisfies conditions (i) and (ii). The subspaces $\{\mathbf{0}\}$ and ${\mathbf{\text{R}}}^3$ are sometimes called the trivial subspaces of ${\mathbf{\text{R}}}^3$ (because the verification that they are subspaces is quite trivial). All subspaces other than $\{\mathbf{0}\}$ and ${\mathbf{\text{R}}}^3$ itself are called proper subspaces of ${\mathbf{\text{R}}}^3$.

Now we want to show that the proper subspaces of ${\mathbf{\text{R}}}^3$ are what we customarily call lines and planes through the origin. Let V be a subspace of ${\mathbf{\text{R}}}^3$ that is neither $\{\mathbf{0}\}$ nor ${\mathbf{\text{R}}}^3$ itself. There are two cases to consider, depending on whether or not V contains two linearly independent vectors.

Case 1: Suppose that V does not contain two linearly independent vectors. If u is a fixed nonzero vector in V, then, by condition (ii) above, every scalar multiple cu is also in V. Conversely, if v is any other vector in V, then u and v are linearly dependent, so it follows that $\mathbf{\text{v}}=c\mathbf{\text{u}}$ for some scalar c. Thus the subspace V is the set of all scalar multiples of the fixed nonzero vector u and is therefore what we call a line through the origin in ${\mathbf{\text{R}}}^3$. (See Fig. 4.1.8.)

Case 2: Suppose that V contains two linearly independent vectors u and v. It then follows from conditions (i) and (ii) that V contains every linear combination $a\mathbf{\text{u}}+b\mathbf{\text{v}}$ of u and v. (See Problem 38.) Conversely, let w be any other vector in V. If u, v, w were linearly independent, then, by Theorem 4, V would be all of ${\mathbf{\text{R}}}^3$. Therefore, u, v, w are linearly dependent, so it follows that there exist scalars a and b such that $\mathbf{\text{w}}=a\mathbf{\text{u}}+b\mathbf{\text{v}}$. (See Problem 40.) Thus the subspace V is the set of all linear combinations $a\mathbf{\text{u}}+b\mathbf{\text{v}}$ of the two linearly independent vectors u and v and is therefore what we call a plane through the origin in ${\mathbf{\text{R}}}^3$. (See Fig. 4.1.9.)

Subspaces of the coordinate plane ${\mathbf{\text{R}}}^2$ are defined similarly—they are the nonempty subsets of ${\mathbf{\text{R}}}^2$ that are closed under addition and multiplication by scalars. In Problem 39 we ask you to show that every proper subspace of ${\mathbf{\text{R}}}^2$ is a line through the origin.

Example 6 illustrates the fact that lines that do not pass through the origin are not subspaces of ${\mathbf{\text{R}}}^2$. Because every subspace must contain the zero vector (per Problem 37), only lines and planes that pass through the origin are subspaces of ${\mathbf{\text{R}}}^3$.