Page 48 - Linear Algebra Done Right

Chapter 2. Finite-Dimensional Vector Spaces
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for some choice of scalars d 1 ,...,d m .
But (u 1 ,...,u m ,w 1 ,...,w k )
is linearly independent, so the last equation implies that all the c’s
(and d’s) equal 0. Thus our original equation involving the a’s, b’s, and
c’s becomes
a 1 u 1 +· · ·+ a m u m + b 1 v 1 +· · ·+ b j v j = 0.
This equation implies that all the a’s and b’s are 0 because the list
(u 1 ,...,u m ,v 1 ,...,v j ) is linearly independent. We now know that all
the a’s, b’s, and c’s equal 0, as desired.

The next proposition shows that dimension meshes well with direct
sums. This result will be useful in later chapters.

Recall that direct sum 2.19 Proposition: Suppose V is finite dimensional and U 1 ,...,U m
is analogous to disjoint are subspaces of V such that
union. Thus 2.19 is
analogous to the 2.20 V = U 1 +· · ·+ U m
statement that if a
and
finite set B is written as
A 1 ∪ ··· ∪ A m and the
2.21 dim V = dim U 1 +· · ·+ dim U m .
sum of the number of
elements in the A’s Then V = U 1 ⊕· · · U m .
equals the number of
elements in B, then the Proof: Choose a basis for each U j . Put these bases together in
union is a disjoint one list, forming a list that spans V (by 2.20) and has length dim V
union. (by 2.21). Thus this list is a basis of V (by 2.16), and in particular it is
linearly independent.
Now suppose that u 1 ∈ U 1 ,...,u m ∈ U m are such that

0 = u 1 + ··· + u m .

We can write each u j as a linear combination of the basis vectors (cho-
sen above) of U j . Substituting these linear combinations into the ex-
pression above, we have written 0 as a linear combination of the basis
of V constructed above. Thus all the scalars used in this linear combina-
tion must be 0. Thus each u j = 0, which proves that V = U 1 ⊕···⊕U m
(by 1.8).