Dimension of a subspace

Theorem 1

Theorem 1. A subspace in an -dimensional vector space is a vector space of dimension .

Proof.

Proof. It is possible to give a deceptively short proof of this theorem that runs as follows. Every set of vectors in is linearly dependent, hence the same is true of ; hence, in particular, the number of elements in each basis of is , Q.E.D.

The trouble with this argument is that we defined dimension by requiring in the first place that there exist a finite basis, and then demanding that this basis contain exactly elements. The proof above shows only that no basis can contain more than elements; it does not show that any basis exists. Once the difficulty is observed, however, it is easy to fill the gap. If , then is -dimensional, and we are done. If contains a non-zero vector , let ( ) be the subspace spanned by . If , then is -dimensional, and we are done. If , let be an element of not contained in , and let be the subspace spanned by and ; and so on. Now we may legitimately employ the argument given above; after no more than steps of this sort, the process reaches an end, since (by Section: Dimension , Theorem 2) we cannot find linearly independent vectors. ◻

∎

The following result is an important consequence of this second and correct proof of Theorem 1.

Theorem 2

Theorem 2. Given any -dimensional subspace in an -dimensional vector space , we can find a basis in so that are in and form, therefore, a basis of .

We shall denote the dimension of a vector space by the symbol . In this notation Theorem 1 asserts that if is a subspace of a finite-dimensional vector space , then .

EXERCISES

Exercise 1.

Exercise 1. If and are finite-dimensional subspaces with the same dimension, and if , then .

Exercise 2.

Exercise 2. If and are subspaces of a vector space , and if every vector in belongs either to or to (or both), then either or (or both).

Exercise 3.

Exercise 3. If , , and are vectors such that , then and span the same subspace as and .

Exercise 4.

Exercise 4. Suppose that and are vectors and is a subspace in a vector space ; let be the subspace spanned by and , and let be the subspace spanned by and . Prove that if is in but not in , then is in .

Exercise 5.

Exercise 5. Suppose that , , and are subspaces of a vector space.

Show that the equation is not necessarily true.
Prove that

Exercise 6.

Can it happen that a non-trivial subspace of a vector space (i.e., a subspace different from both and ) has a unique complement?
If is an -dimensional subspace in an -dimensional vector space, then every complement of has dimension .

Exercise 7.

Show that if both and are three-dimensional subspaces of a five-dimensional vector space, then and are not disjoint.
If and are finite-dimensional subspaces of a vector space, then

Exercise 8.

Exercise 8. A polynomial is called even if identically in (see Section: Subspaces , (3)), and it is called odd if .

(a) Both the class of even polynomials and the class of odd polynomials are subspaces of the space of all (complex) polynomials.

(b) Prove that and are each other’s complements.

EXERCISES

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