Page 44 - Linear Algebra Done Right

Chapter 2. Finite-Dimensional Vector Spaces
30
Step j
If w j is in the span of B, leave B unchanged. If w j is not in the
span of B, extend B by adjoining w j to it.
After each step, B is still linearly independent because otherwise the
linear dependence lemma (2.4) would give a contradiction (recall that
(v 1 ,...,v m ) is linearly independent and any w j that is adjoined to B is
not in the span of the previous vectors in B). After step n, the span of
B includes all the w’s. Thus the B obtained after step n spans V and
hence is a basis of V.

As a nice application of the theorem above, we now show that ev-
ery subspace of a finite-dimensional vector space can be paired with
another subspace to form a direct sum of the whole space.

Using the same basic 2.13 Proposition: Suppose V is finite dimensional and U is a sub-
ideas but considerably space of V. Then there is a subspace W of V such that V = U ⊕ W.
more advanced tools,
this proposition can be Proof: Because V is finite dimensional, so is U (see 2.7). Thus
proved without the there is a basis (u 1 ,...,u m ) of U (see 2.11). Of course (u 1 ,...,u m )
hypothesis that V is is a linearly independent list of vectors in V, and thus it can be ex-
finite dimensional. tended to a basis (u 1 ,...,u m ,w 1 ,...,w n ) of V (see 2.12). Let W =
span(w 1 ,...,w n ).
To prove that V = U ⊕ W, we need to show that

V = U + W and U ∩ W ={0};
see 1.9. To prove the first equation, suppose that v ∈ V. Then,
because the list (u 1 ,...,u m ,w 1 ,...,w n ) spans V, there exist scalars
a 1 ,...,a m ,b 1 ,...,b n ∈ F such that
v = a 1 u 1 +· · ·+ a m u m + b 1 w 1 +· · ·+ b n w n .

u w
In other words, we have v = u+w, where u ∈ U and w ∈ W are defined
as above. Thus v ∈ U + W, completing the proof that V = U + W.
To show that U ∩ W ={0}, suppose v ∈ U ∩ W. Then there exist
scalars a 1 ,...,a m ,b 1 ,...,b n ∈ F such that
v = a 1 u 1 + ··· + a m u m = b 1 w 1 +· · ·+ b n w n .

Thus