Page 46 - Linear Algebra Done Right

Chapter 2. Finite-Dimensional Vector Spaces
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2.15
Proposition:
If V is finite dimensional and U is a subspace
of V, then dim U ≤ dim V.
Proof: Suppose that V is finite dimensional and U is a subspace
of V. Any basis of U is a linearly independent list of vectors in V and
thus can be extended to a basis of V (by 2.12). Hence the length of a
basis of U is less than or equal to the length of a basis of V.
The real vector space To check that a list of vectors in V is a basis of V, we must, according
2
R has dimension 2; to the definition, show that the list in question satisfies two properties:
the complex vector it must be linearly independent and it must span V. The next two
space C has results show that if the list in question has the right length, then we
dimension 1. As sets, need only check that it satisfies one of the required two properties.
2
R can be identified We begin by proving that every spanning list with the right length is a
with C (and addition is basis.
the same on both
spaces, as is scalar 2.16 Proposition: If V is finite dimensional, then every spanning
multiplication by real list of vectors in V with length dim V is a basis of V.
numbers). Thus when
we talk about the Proof: Suppose dim V = n and (v 1 ,...,v n ) spans V. The list
dimension of a vector (v 1 ,...,v n ) can be reduced to a basis of V (by 2.10). However, every
space, the role played basis of V has length n, so in this case the reduction must be the trivial
by the choice of F one, meaning that no elements are deleted from (v 1 ,...,v n ). In other
cannot be neglected. words, (v 1 ,...,v n ) is a basis of V, as desired.

Now we prove that linear independence alone is enough to ensure
that a list with the right length is a basis.

2.17 Proposition: If V is finite dimensional, then every linearly
independent list of vectors in V with length dim V is a basis of V.

Proof: Suppose dim V = n and (v 1 ,...,v n ) is linearly independent
in V. The list (v 1 ,...,v n ) can be extended to a basis of V (by 2.12). How-
ever, every basis of V has length n, so in this case the extension must be
the trivial one, meaning that no elements are adjoined to (v 1 ,...,v n ).
In other words, (v 1 ,...,v n ) is a basis of V, as desired.

As an example of how the last proposition can be applied, consider
2
the list (5, 7), (4, 3) . This list of two vectors in F is obviously linearly
independent (because neither vector is a scalar multiple of the other).