Subspaces
Definition 1.5.1 A subspace of a vector space V is a subset S of V that is a vector space in its own right under the operations obtained by restricting the operations of V to S. We use the notation S ≤ V to indicate that S is a subspace of V and S < V to indicate that S is a proper subspace of V, that is, S ≤ V but S ≠ V. The zero subspace of V is {0}. Then S is a subspace of V if and only if the following conditions hold:
If u and v are in S, then u + v is in S.
If u is in S and λ is a scalar, then λu is in S.
An equivalent definition in terms of groups is the following
Definition 1.5.2 Let V be a vector space over a field F. A non-empty subset U of V is called a subspace of V iff
(U, +) is a subgroup of (V , +);
for α ∈ F and u ∈ U, αu ∈ U.
Clearly, for any vector space V, {0} and V are subspaces of V, called trivial subspaces. Any other subspace of V (if it exists) is called a proper subspace of V.
Example 1.5.3 If V is a vector space, then V is clearly a subspace of itself. The set {0}, consisting of only the zero vector, is also a subspace of V, called the zero subspace. To show this, we simply note that the two closure conditions are satisfied:
0 + 0 = 0 and λ0 = 0 for any scalar λ
The subspaces {0} and V are called the trivial subspaces of V. ■
Dimension of a Subspace
It is obvious that the dimension of a subspace U of a vector space V cannot be greater than the dimension of the entire space V.
Theorem 1.5.4 If the dimension of a subspace U of a vector space V is equal to the dimension of V, then the subspace U is equal to all of V.
Proof Suppose dim U = dim V = n. Then in U one could find n linearly independent vectors x1, ... ,xn. If U ≠ V, then in V there would be some vector x ≠ U. Since dim V = n, it follows that any n + 1 vectors in this space are linearly dependent. In particular, the vectors x1, ... , xn, x are linearly dependent. That is, there is a relationship
α1 x1 + ··· + αnxn + αx = 0
with not all coefficients equal to zero. If we had α = 0, then this would yield the linear dependence of the vectors x1, ... , xn, which are linearly independent by assumption. This means that α ≠ 0 and x = β1x1 + ··· + βnxn, βi = −α−1αi, from which it follows that x is a linear combination of the vectors x1, ... , xn. It clearly follows from the definition of a subspace that a linear combination of vectors in U is itself a vector in U. Hence we have x ∈ U, and U = V. □