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Unformatted text preview: 372 CHAPTER 3. REALVALUED FUNCTIONS: DIFFERENTIATION
Diﬀerentiating with respect to x1 , this gives
∂f
= 2a11 x1 + 2a12 x2 + 2a13 x3
∂x1
= 2(a11 x1 + a12 x2 + a13 x3 ).
Note that the quantity in parentheses is exactly the product of the ﬁrst
→
→
row of A = [Q] with [− ], or equivalently the ﬁrst entry of A [− ]. For
x
x
→
− for the vector
convenience, we will abuse notation, and write simply A x
→
whose coordinate column is A times the coordinate column of − :
x
→
→
[A− ] = A [− ] .
x
x
You should check that the other two partials of f are the other coordinates
→
of A− , so
x
→→
−−
→
∇ f ( u ) = 2A− .
u
If we also recall that the dot product of two vectors can be written in
terms of their coordinate columns as
→→
− · − = [− ]T [− ]
→→
xy
x
y
→
→
then the matrix form of f (− ) = Q(− ) becomes
x
x
→
→→
f (− ) = − · A− ;
x
x
x
we separate out this calculation as
Remark 3.9.1. The gradient of a function of the form
→
→→
f (− ) = − · A−
x
x
x
is →→
−−
→
∇ f ( x ) = 2A− .
x Note that, while our calculation was for a 3 × 3 matrix, the analogous
result holds for a 2 × 2 matrix as well.
Now, the Lagrange multiplier condition for extrema of f on S 2 becomes
→
→
A− = λ− .
u
u (3.31) →
→
Geometrically, this means that A− and − have the same direction (up to
u
u
reversal, or possibly squashing to zero). Such situations come up often in
→
problems involving matrices; we call a nonzero vector − which satisﬁes
u ...
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Spring '08 term at University of Florida.
 Spring '08
 ALL
 Calculus

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