Linear Algebra, 2020a

126 Chapter Two. Vector Spaces
so that ⃗v 1 = c 1,1
⃗β 1 + ···+ c n,1
⃗β n , etc. Then a 1 ⃗v 1 + ···+ a k ⃗v k = ⃗0 is equivalent
to these.
⃗0 = a 1 · (c 1,1
⃗β 1 + ···+ c n,1
⃗β n )+···+ a k · (c 1,k
⃗β 1 + ···+ c n,k
⃗β n )
=(a 1 c 1,1 + ···+ a k c 1,k ) · ⃗β 1 + ···+(a 1 c n,1 + ···+ a k c n,k ) · ⃗β n
Obviously the bottom equation is true if the coefficients are zero. But, because
B is a basis, Theorem 1.12 says that the bottom equation is true if and only if
the coefficients are zero. So the relation is equivalent to this.
a 1 c 1,1 + ···+ a k c 1,k = 0
.
a 1 c n,1 + ···+ a k c n,k = 0
This is the equivalent recast into column vectors.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
c 1,1
c 1,k 0
⎜
a 1 ⎝
⎟ ⎜ .
. ⎠ + ···+ a k ⎝
⎟ ⎜
. ⎠ = ⎝
⎟
. ⎠
c n,1 c n,k 0
Note that not only does a relationship hold for one set if and only if it holds for
the other, but it is the same relationship — the a i are the same. QED
1.19 Example Example 1.14 finds the representation of x + x 2 ∈ P 3 with respect
to B = 〈1, 2x, 2x 2 ,2x 3 〉.
⎛ ⎞
0
Rep B (x + x 2 1/2
)= ⎜ ⎟
⎝1/2⎠
0
This relationship
2 · (x + x 2 )−1 · (2x)−2 · (x 2 )=0 + 0x + 0x 2 + 0x 3
is represented by this one.
⎛ ⎞ ⎛ ⎞
0 0
−2·⎛ ⎞ ⎛ ⎞
0 0
2·Rep B (x+x 2 )−Rep B (2x)−2·Rep B (x 2 1/2
)=2· ⎜ ⎟
⎝1/2⎠ − 1
0
⎜ ⎟ ⎜ ⎟
⎝0⎠ ⎝1/2⎠ = 0
⎜ ⎟
⎝0⎠
0 0 0 0
Our main use of representations will come later but the definition appears
here because the fact that every vector is a linear combination of basis vectors in
a unique way is a crucial property of bases, and also to help make a point. For
calculation of coordinates among other things, we shall restrict our attention
to spaces with bases having only finitely many elements. That will start in the
next subsection.
B