Theorem 1. A subspace \(\mathcal{M}\) in an \(n\) -dimensional vector space \(\mathcal{V}\) is a vector space of dimension \(\leq n\) .
Proof. It is possible to give a deceptively short proof of this theorem that runs as follows. Every set of \(n+1\) vectors in \(\mathcal{V}\) is linearly dependent, hence the same is true of \(\mathcal{M}\) ; hence, in particular, the number of elements in each basis of \(\mathcal{M}\) 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 \(\mathcal{M}=\mathcal{O}\) , then \(\mathcal{M}\) is \(0\) -dimensional, and we are done. If \(\mathcal{M}\) contains a non-zero vector \(x_{1}\) , let \(\mathcal{M}_{1}\) ( \(\subset \mathcal{M}\) ) be the subspace spanned by \(x_{1}\) . If \(\mathcal{M}=\mathcal{M}_{1}\) , then \(\mathcal{M}\) is \(1\) -dimensional, and we are done. If \(\mathcal{M} \neq \mathcal{M}_{1}\) , let \(x_2\) be an element of \(\mathcal{M}\) not contained in \(\mathcal{M}_1\) , and let \(\mathcal{M}_2\) 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. Given any \(m\) -dimensional subspace \(\mathcal{M}\) in an \(n\) -dimensional vector space \(\mathcal{V}\) , we can find a basis \(\{x_{1}, \ldots, x_{m}, x_{m+1}, \ldots, x_{n}\}\) in \(\mathcal{V}\) so that \(x_{1}, \ldots, x_{m}\) are in \(\mathcal{M}\) and form, therefore, a basis of \(\mathcal{M}\) .
We shall denote the dimension of a vector space \(\mathcal{V}\) by the symbol \(\operatorname{dim} \mathcal{V}\) . In this notation Theorem 1 asserts that if \(\mathcal{M}\) is a subspace of a finite-dimensional vector space \(\mathcal{V}\) , then \(\operatorname{dim} \mathcal{M} \leq \operatorname{dim} \mathcal{V}\) .
EXERCISES
Exercise 1. If \(\mathcal{M}\) and \(\mathcal{N}\) are finite-dimensional subspaces with the same dimension, and if \(\mathcal{M} \subset \mathcal{N}\) , then \(\mathcal{M}=\mathcal{N}\) .
Exercise 2. If \(\mathcal{M}\) and \(\mathcal{N}\) are subspaces of a vector space \(\mathcal{V}\) , and if every vector in \(\mathcal{V}\) belongs either to \(\mathcal{M}\) or to \(\mathcal{N}\) (or both), then either \(\mathcal{M}=\mathcal{V}\) or \(\mathcal{N}=\mathcal{V}\) (or both).
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. Suppose that \(x\) and \(y\) are vectors and \(\mathcal{M}\) is a subspace in a vector space \(\mathcal{V}\) ; let \(\mathcal{H}\) be the subspace spanned by \(\mathcal{M}\) and \(x\) , and let \(\mathcal{K}\) be the subspace spanned by \(\mathcal{M}\) and \(y\) . Prove that if \(y\) is in \(\mathcal{H}\) but not in \(\mathcal{M}\) , then \(x\) is in \(\mathcal{K}\) .
Exercise 5. Suppose that \(\mathcal{L}\) , \(\mathcal{M}\) , and \(\mathcal{N}\) are subspaces of a vector space.
- Show that the equation \[\mathcal{L} \cap(\mathcal{M}+\mathcal{N})=(\mathcal{L} \cap \mathcal{M})+(\mathcal{L} \cap \mathcal{N})\] is not necessarily true.
- Prove that \[\mathcal{L} \cap\big(\mathcal{M}+(\mathcal{L} \cap \mathcal{N})\big)=(\mathcal{L} \cap \mathcal{M})+(\mathcal{L} \cap \mathcal{N})\]
Exercise 6.
- Can it happen that a non-trivial subspace of a vector space \(\mathcal{V}\) (i.e., a subspace different from both \(\mathcal{O}\) and \(\mathcal{V}\) ) has a unique complement?
- If \(\mathcal{M}\) is an \(m\) -dimensional subspace in an \(n\) -dimensional vector space, then every complement of \(\mathcal{M}\) has dimension \(n-m\) .
Exercise 7.
- Show that if both \(\mathcal{M}\) and \(\mathcal{N}\) are three-dimensional subspaces of a five-dimensional vector space, then \(\mathcal{M}\) and \(\mathcal{N}\) are not disjoint.
- If \(\mathcal{M}\) and \(\mathcal{N}\) are finite-dimensional subspaces of a vector space, then \[\operatorname{dim} \mathcal{M}+\operatorname{dim} \mathcal{N}=\operatorname{dim}(\mathcal{M}+\mathcal{N})+\operatorname{dim}(\mathcal{M} \cap \mathcal{N}).\]
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 \(\mathcal{M}\) of even polynomials and the class \(\mathcal{N}\) of odd polynomials are subspaces of the space \(\mathcal{P}\) of all (complex) polynomials.
(b) Prove that \(\mathcal{M}\) and \(\mathcal{N}\) are each other’s complements.