Page 418 - 35Linear Algebra

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Essentially, this is already our change of basis formula, but lets play around
and put it in our notations. First we can write this as a matrix

s λ 2λ x
= .
f 1 1 y
We can easily invert this to get
1
x − 2 s
= λ .
1
y −1 f
λ
Putting this in the engineer’s formula for ~x gives
1
− 2 s s
λ 1
~x = ~e 1 ~e 2 1 −1 f = − λ ~e 1 − ~e 2 2~e 1 − 2~e 2 f .
λ
Comparing to the nutritionist’s formula for the same object ~x we learn that
1
~
~
f 1 = − ~e 1 − ~e 2 and f 2 = 2~e 1 − 2~e 2 .
λ
Rearranging these equation we find the change of base matrix P from the engi-
neer’s basis to the nutritionist’s basis:

− 1 2
~ ~ λ
f 1 f 2 = ~e 1 ~e 2 1 −1 =: ~e 1 ~e 2 P .
λ
We can also go the other direction, changing from the nutritionist’s basis to
the engineer’s basis

~
~
~
λ 2λ
~
~e 1 ~e 2 = f 1 f 2 =: f 1 f 2 Q .
1 1
Of course, we must have
Q = P −1 ,
(which is in fact how we constructed P in the first place).
Finally, lets consider the very first linear systems problem, where you
were given that there were 27 pieces of fruit in total and twice as many oranges
as apples. In equations this says just
x + y = 27 and 2x − y = 0 .

But we can also write this as a matrix system

MX = V

where

1 1 x 0
M := , X := V := .
2 −1 y 27
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