Linear Algebra

Section III. Basis and Dimension 135
and vice versa.
For (1) =⇒ (2), assume that all decompositions are unique. We will show
⌢ ⌢
that B1 ··· Bk spans the space and is linearly independent. It spans the
space because the assumption that V = W1 + ··· + Wk means that every �v
canbeexpressedas�v = �w1 + ···+ �wk, which translates by equation (∗) toan
expression of �v as a linear combination of the � β’s from the concatenation. For
linear independence, consider this linear relationship.
�0 =c1,1 � �βk,nk
β1,1 + ···+ ck,nk
Regroup as in (∗) (that is, take d1, ... , dk to be 1 and move from bottom to
top) to get the decomposition �0 = �w1 + ···+ �wk. Because of the assumption
that decompositions are unique, and because the zero vector obviously has the
decomposition �0 =�0+···+�0, we now have that each �wi is the zero vector. This
means that ci,1 � �βi,ni βi,1 + ···+ ci,ni = �0. Thus, since each Bi is a basis, we have
the desired conclusion that all of the c’s are zero.
⌢ ⌢
For (2) =⇒ (3), assume that B1 ··· Bk is a basis for the space. Consider
a linear relationship among nonzero vectors from different Wi’s,
�0 = ···+ di �wi + ···
in order to show that it is trivial. (The relationship is written in this way
because we are considering a combination of nonzero vectors from only some of
the Wi’s; for instance, there might not be a �w1 in this combination.) As in (∗),
�0 = ···+di(ci,1 � �βi,ni βi,1+···+ci,ni )+··· = ···+dici,1· � βi,1+···+dici,ni ·� βi,ni +···
⌢ ⌢
and the linear independence of B1 ··· Bk gives that each coefficient dici,j is
zero. Now, �wi is a nonzero vector, so at least one of the ci,j’s is zero, and thus
di is zero. This holds for each di, and therefore the linear relationship is trivial.
Finally, for (3) =⇒ (1), assume that, among nonzero vectors from different
Wi’s, any linear relationship is trivial. Consider two decompositions of a vector
�v = �w1 + ···+ �wk and �v = �u1 + ···+ �uk in order to show that the two are the
same. We have
�0 =(�w1 + ···+ �wk) − (�u1 + ···+ �uk) =(�w1 − �u1)+···+(�wk − �uk)
which violates the assumption unless each �wi − �ui is the zero vector. Hence,
decompositions are unique. QED
4.9 Definition A collection of subspaces {W1,... ,Wk} is independent if no
nonzero vector from any Wi is a linear combination of vectors from the other
subspaces W1,...,Wi−1,Wi+1,...,Wk.
4.10 Definition A vector space V is the direct sum (or internal direct sum)
of its subspaces W1,...,Wk if V = W1 + W2 + ··· + Wk and the collection
{W1,...,Wk} is independent. We write V = W1 ⊕ W2 ⊕ ...⊕ Wk.
4.11 Example The benchmark model fits: R 3 = x-axis ⊕ y-axis ⊕ z-axis.