CHAPTER II DIMENSION In the present chapter we investigate ...

Applying T to this identity, we have T (au + bv) = 0, or aT u + bT v = 0. From
T u = 2u and T v = 3v we get
2a u + 3b v = 0. (1.2.4)
Multiply (1.2.3) by 2 and then substract (1.2.4), we get bv = 0. Since v is nonzero, we
must have b = 0. Thus (1.2.3) becomes au = 0. Since u is nonzero, we must have a = 0.
We have shown a = b = 0. Hence u, v are linearly independent.
Example 1.2.4. Let c1, c2, . . . , cn, cn+ 1 be n + 1 distinct complex numbers and con-
) for k = 0, 1, 2, . . . , n:
sider vectors in C n+ 1 : vk = (c k 1 , ck 2 , . . . , ck n , ck n+ 1
v0 = (1, 1, 1, . . . , 1, 1)
v1 = (c1, c2, . . . , cn, cn+ 1)
v2 = (c 2 1 , c2 2 , . . . , c2 n , c2 n+ 1 )
.
vn = (c n 1 , cn 2 , . . . , cn n , cn n+ 1 )
Prove that v0, v1, v2, . . . , vn are linearly independent.
Proof: Assume a0v0 + a1v1 + · · · + anvn = 0. For each j with 1 ≤ j ≤ n + 1, write
out the jth component of this vector identity:
a0 + a1cj + a2c 2 j + · · · + anc n j
This shows that the polynomial a0 +a1x+a2x 2 +· · ·+anx n has n+1 roots c1, c2, . . . , cn+ 1.
But a polynomial of degree at most n cannot have n + 1 roots, unless that polynomial is
the zero polynomial. Therefore a0, a1, . . . , an must vanish. This shows v0, v1, v2, . . . , vn
are linearly independent.
1.3. Given a finite set of vectors v1, v2, . . . , vr in V , we can construct a subspace
containing these vectors by collecting all linear combinations of these vectors. Here, by a
linear combination of v1, v2, . . . , vr we mean a vector of the form
α1v1 + α2v2 + · · · · · · + αrvr
= 0.
(1.3.1)
for certain scalars α1, α2, . . . , αr. A quick example: 4 + 2x − x 2 is a linear combination of
2 + 3x and 4x + x 2 because 4 + 2x − x 2 = 2(2 + 3x) + (−1)(4x + x 2 ), which can be easily
checked. (For the moment let us not worry about how to figure out this identity.) One
can think of a linear combination of the form (1.3.1) as a “combo soup” with vk as its
kth ingredient and ak as the amount of this ingredient (1 ≤ k ≤ r).
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