“real: chapter_12” — 2011/5/23 — 1:04 — page 40 — #40
12-40 Real Analysis
for some real 2 ×2 matrix A. Using the standard basis (1, 0) and (0, 1)
of the vector space
R
2
over R, we can easily see that the real linear
map satisfies
L(h, k) = L(h(1, 0) + k(0, 1)) = hL(1, 0) + kL(0, 1)
for all (h, k) ∈
R
2
.IfweletL(1, 0) = (α, β) and L(0, 1) = (γ , δ), then
L(h, k) = (hα + kγ , hβ +kδ) =
αγ
βδ
h
k
T
.
Thus the required A can be taken as
αγ
βδ
and one part of our theorem is proved. To prove the converse, we shall
assume
f (x + h, y + k) = f (x, y) + L(h, k) + R(h, k), (h, k) ∈ N
and define
L(h, k) =
A
h
k
T
from R
2
to R
2
. We shall verify that this L is real linear, so that by
definition, f will be real differentiable. Now
L((h, k) + (u, v)) = L(h + u, k + v) =
A