Dimension of a subspace

Theorem 1

Theorem 1. A subspace $M$ in an $n$ -dimensional vector space $V$ is a vector space of dimension $\leq n$ .

Proof.

Proof. It is possible to give a deceptively short proof of this theorem that runs as follows. Every set of $n + 1$ vectors in $𝒱$ is linearly dependent, hence the same is true of $ℳ$ ; hence, in particular, the number of elements in each basis of $ℳ$ is $\leq n$ , Q.E.D.

The trouble with this argument is that we defined dimension $n$ by requiring in the first place that there exist a finite basis, and then demanding that this basis contain exactly $n$ elements. The proof above shows only that no basis can contain more than $n$ elements; it does not show that any basis exists. Once the difficulty is observed, however, it is easy to fill the gap. If $ℳ = 𝒪$ , then $ℳ$ is $0$ -dimensional, and we are done. If $ℳ$ contains a non-zero vector $x_{1}$ , let $ℳ_{1}$ ( $\subset ℳ$ ) be the subspace spanned by $x_{1}$ . If $ℳ = ℳ_{1}$ , then $ℳ$ is $1$ -dimensional, and we are done. If $ℳ \neq ℳ_{1}$ , let $x_{2}$ be an element of $ℳ$ not contained in $ℳ_{1}$ , and let $ℳ_{2}$ be the subspace spanned by $x_{1}$ and $x_{2}$ ; and so on. Now we may legitimately employ the argument given above; after no more than $n$ steps of this sort, the process reaches an end, since (by Section: Dimension , Theorem 2) we cannot find $n + 1$ 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 $m$ -dimensional subspace $ℳ$ in an $n$ -dimensional vector space $𝒱$ , we can find a basis ${x_{1}, \dots, x_{m}, x_{m + 1}, \dots, x_{n}}$ in $𝒱$ so that $x_{1}, \dots, x_{m}$ are in $ℳ$ and form, therefore, a basis of $ℳ$ .

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

EXERCISES

Exercise 1.

Exercise 1. If $ℳ$ and $𝒩$ are finite-dimensional subspaces with the same dimension, and if $ℳ \subset 𝒩$ , then $ℳ = 𝒩$ .

Exercise 2.

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

Exercise 3.

Exercise 3. If $x$ , $y$ , and $z$ are vectors such that $x + y + z = 0$ , then $x$ and $y$ span the same subspace as $y$ and $z$ .

Exercise 4.

Exercise 4. Suppose that $x$ and $y$ are vectors and $ℳ$ is a subspace in a vector space $𝒱$ ; let $ℋ$ be the subspace spanned by $ℳ$ and $x$ , and let $𝒦$ be the subspace spanned by $ℳ$ and $y$ . Prove that if $y$ is in $ℋ$ but not in $ℳ$ , then $x$ is in $𝒦$ .

Exercise 5.

Exercise 5. Suppose that $ℒ$ , $ℳ$ , and $𝒩$ are subspaces of a vector space.

Show that the equation $ℒ \cap (ℳ + 𝒩) = (ℒ \cap ℳ) + (ℒ \cap 𝒩)$ is not necessarily true.
Prove that $ℒ \cap (ℳ + (ℒ \cap 𝒩)) = (ℒ \cap ℳ) + (ℒ \cap 𝒩)$

Exercise 6.

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

Exercise 7.

Show that if both $ℳ$ and $𝒩$ are three-dimensional subspaces of a five-dimensional vector space, then $ℳ$ and $𝒩$ are not disjoint.
If $ℳ$ and $𝒩$ are finite-dimensional subspaces of a vector space, then $\dim ℳ + \dim 𝒩 = \dim (ℳ + 𝒩) + \dim (ℳ \cap 𝒩) .$

Exercise 8.

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

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

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

EXERCISES

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