Consider the collection of vectors

The endpoints of all such vectors lie on the line *y* = 3 *x* in the *x‐y* plane. Now, choose any two vectors from *V*, say, **u** = (1, 3) and **v** = (‐2, ‐6). Note that the sum of **u** and **v**,

is also a vector in *V*, because its second component is three times the first. In fact, it can be easily shown that the sum of *any* two vectors in *V* will produce a vector that again lies in *V*. The set *V* is therefore said to be **closed under addition**. Next, consider a scalar multiple of **u**, say,

It, too, is in *V*. In fact, *every* scalar multiple of any vector in *V* is itself an element of *V*. The set *V* is therefore said to be **closed under scalar multiplication**.

Thus, the elements in *V* enjoy the following two properties:

Closure under addition:

The sum of any two elements in *V* is an element of *V*.

Closure under scalar multiplication:

Every scalar multiple of an element in *V* is an element of *V*.

Any subset of **R** ^{n }that satisfies these two properties—with the usual operations of addition and scalar multiplication—is called a **subspace of R**^{n }or a **Euclidean vector space**. The set *V* = {(*x*, 3 *x*): *x* ∈ **R**} is a Euclidean vector space, a subspace of **R**^{2}.

**Example 1**: Is the following set a subspace of **R**^{2}?

To establish that *A* is a subspace of **R**^{2}, it must be shown that *A* is closed under addition and scalar multiplication. If a counterexample to even one of these properties can be found, then the set is not a subspace. In the present case, it is very easy to find such a counterexample. For instance, both **u** = (1, 4) and **v** = (2, 7) are in *A*, but their sum, **u** + **v** = (3, 11), is not. In order for a vector **v** = (*v* _{1}, *v* _{2} to be in *A*, the second component (*v* _{2}) must be 1 more than three times the first component (*v* _{1}). Since 11 ≠ 3(3) + 1, (3, 11) ∉ *A*. Therefore, the set *A* is not closed under addition, so *A* cannot be a subspace. [You could also show that this particular set is not a subspace of **R** ^{2} by exhibiting a counterexample to closure under scalar multiplication. For example, although **u** = (1, 4) is in *A*, the scalar multiple 2 **u** = (2, 8) is not.]

**Example 2**: Is the following set a subspace of **R**^{3}?

In order for a sub set of **R** ^{3} to be a sub space of **R** ^{3}, both closure properties (1) and (2) must be satisfied. However, note that while **u** = (1, 1, 1) and **v** = (2, 4, 8) are both in *B*, their sum, (3, 5, 9), clearly is not. Since *B* is not closed under addition, *B* is not a subspace of **R** ^{3}.

**Example 3**: Is the following set a subspace of **R**^{4}?

For a 4‐vector to be in *C*, exactly two conditions must be satisfied: Namely, its second component must be zero, and its fourth component must be −5 times the first. Choosing particular vectors in *C* and checking closure under addition and scalar multiplication would lead you to conjecture that *C* is indeed a subspace. However, no matter how many specific examples you provide showing that the closure properties are satisfied, the fact that *C* is a subspace is established only when a general proof is given. So let **u** = (*u*_{1}, 0, *u*_{3}, −5 *u*_{1}) and **v** = (*v*_{1}, 0, *v*_{3}, −5*v*_{1}) be arbitrary vectors in *C*. Then their sum,

satisfies the conditions for membership in *C*, verifying closure under addition. Finally, if *k* is a scalar, then

is in *C*, establishing closure under scalar multiplication. This proves that *C* is a subspace of **R**^{4}.

**Example 4**: Show that if *V* is a subspace of **R**^{ n}, then *V* must contain the zero vector.

First, choose any vector **v** in *V*. Since *V* is a subspace, it must be closed under scalar multiplication. By selecting 0 as the scalar, the vector 0 **v**, which equals **0**, must be in *V*. [Another method proceeds like this: If **v** is in *V*, then the scalar multiple (−1) **v** = − **v** must also be in *V*. But then the sum of these two vectors, **v** + (− **v**) = **0**, mnust be in *V*, since *V* is closed under addition.]

This result can provide a quick way to conclude that a particular set is not a Euclidean space. *If the set does not contain the zero vector, then it cannot be a subspace*. For example, the set *A* in Example 1 above could not be a subspace of **R** ^{2} because it does not contain the vector **0** = (0, 0). It is important to realize that containing the zero vector is a *necessary* condition for a set to be a Euclidean space, not a *sufficient* one. That is, just because a set contains the zero vector does not guarantee that it is a Euclidean space (for example, consider the set *B* in Example 2); the guarantee is that if the set does *not* contain **0**, then it is *not* a Euclidean vector space.

As always, the distinction between vectors and points can be blurred, and sets consisting of points in **R**^{n} can be considered for classification as subspaces.

**Example 5**: Is the following set a subspace of **R**^{2}?

As illustrated in Figure , this set consists of all points in the first quadrant, including the points (*x*, 0) on the *x* axis with *x* > 0 and the points (0, *y*) on the *y* axis with *y* > 0:

**Figure 1**

The set *D* is closed under addition since the sum of nonnegative numbers is nonnegative. That is, if (*x*_{1}, *y*_{1}) and (*x*_{2}, *y*_{2}) are in *D*, then *x*_{1}, *x*_{2}, *y*_{1}, and *y*_{2} are all greater than or equal to 0, so both sums *x*_{1} + *x*_{2} and *y*_{1} + *y*_{2} are greater than or equal to 0. This implies that

However, *D* is not closed under scalar multiplication. If *x* and *y* are both positive, then ( *x, y*) is in *D*, but for any negative scalar *k*,

since *kx* < 0 (and *ky* < 0). Therefore, *D* is not a subspace of **R** ^{2}.

**Example 6**: Is the following set a subspace of **R** ^{2}?

As illustrated in Figure , this set consists of all points in the first and third quadrants, including the axes:

**Figure 2**

The set *E* is closed under scalar multiplication, since if *k* is any scalar, then *k(x, y)* = ( *kx, ky*) is in *E*. The proof of this last statement follows immediately from the condition for membership in *E*. A point is in *E* if the product of its two coordinates is nonnegative. Since *k* ^{2} > 0 for any real *k*,

However, although *E* is closed under scalar multiplication, it is not closed under addition. For example, although **u** = (4, 1) and **v** = (−2, −6) are both in *E*, their sum, (2, −5), is not. Thus, *E* is not a subspace of **R** ^{2}.

**Example 7**: Does the plane *P* given by the equation 2 *x* + *y* − 3 *z* = 0 form a subspace of **R** ^{3}?

One way to characterize *P* is to solve the given equation for *y*,

and write

If **p** _{1} = ( *x* _{1}, 3 *z* _{1} − 2 *x* _{1}, *z* _{1}) and **p** _{2} = ( *x* _{2}, 3 *z* _{2} − 2 *x* _{2}, *z* _{2}) are points in *P*, then their sum,

is also in *P*, so *P* is closed under addition. Furthermore, if **p** = ( *x*, 3 *z* − 2 *x, z*) is a point in *P*, then any scalar multiple,

is also in *P*, so *P* is also closed under scalar multiplication. Therefore, *P* does indeed form a subspace of **R** ^{3}. Note that *P* contains the origin. By contrast, the plane 2 *x* + *y* − 3 *z* = 1, although parallel to *P*, is *not* a subspace of **R** ^{3} because it does not contain (0, 0, 0); recall Example 4 above. In fact, a plane in **R** ^{3} is a subspace of **R** ^{3} if and only if it contains the origin.