Linear algebra
Theorem 2.3 The span of a set of vectors in a vector space V is a subspace of V
2.11 Adjoint operators
Each linear operator has a related operator, called its adjoint, which has important properties.
Defi nition 2.21 Let V be a vector space with an inner product (u, v) defi ned for all u and v in V. Let L be a linear operator whose domain is D, a subspace
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of V, and whose range is R, also a subspace of V. The adjoint of L with respect to the defi ned inner product, written L*, is an operator whose domain is R and whose range is D, such that
Lu v, ( ,u L v* )
(a) Determine A* with respect to the standard inner product on Rn.
(b) Under what conditions is A self-adjoint with respect to the standard inner product on Rn?
(c) Determine A* with respect to an inner product on Rn defi ned by
( , )u v =2u v1 1+u v2 2+u v3 3+ +… u vn n (b) Solution (a) Let u and v be arbitrary vectors in Rn. Then by defi nition of the standard inner product for Rn,
Au v, Au v rep-resentation of the form
A*
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Then
Interchanging the names of the indices in Equation e leads to
( ,u A v* )= ,*
=
=
∑
∑
jn1 in1 a u vj i j j (f)For the expression in Equation c to be equal to the expression in Equation f for all possible u and v, it is required that
a a i n
Thus the adjoint of A with respect to the standard inner product on Rn is the matrix obtained by interchanging the rows and columns of A. Such a matrix is called the transpose matrix, AT. Thus A* = AT.
(b) From Equation g, it is clear that A is self-adjoint with respect to the standard inner product if
a a i n
Such a matrix, whose columns can be interchanged with its rows without changing the matrix, is called a symmetric matrix. Thus an n × n matrix A is self-adjoint with respect to the standard inner product for Rn if and only if A is a symmetric matrix.
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Following along the same lines as part (a), and using the inner product defi ned in Equation b, leads to
Au v, Au Au vi
Assuming A* in the form of Equation d leads to
( , * ) * *
Renaming the indices in Equation k and rearranging leads to
u A v, * ,*
,*
( )
= + += =
∑
in1 2a1iv ui 1∑
jn2 u aj j1v1∑
i=n2∑
j=n1 u ai j ,,i*vj (l)Equation j and Equation l are identical for all possible u and v if and only if
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For example, a pair of adjoints with respect to the inner product of Equation
Part (a) of Example 2.29 illustrates that a symmetric n × n matrix is self-adjoint with respect to the standard inner product for Rn. Thus the matrices developed from the stress tensor in Example 2.25 and the stiffness matrix of Example 2.26 are self-adjoint with respect to the standard inner product.
Example 2.30 The nondimensional differential equation governing the defl ection, w(x), of a non-uniform beam due to a distributed load per unit length, f(x), is
where α(x) is a function describing the variation of material and geomet-ric parameters across the span of the beam. The boundary conditions are dependent on the end constraints. For a beam fi xed at x = 0 and free at x = 1, the appropriate boundary conditions are
w( )0 = (b1)0 that if g(x) is in D, then g(x) satisfi es the boundary conditions of Equation b.
Consider the range of V to be the same as its domain. The standard inner product on V is
( ( ), ( ))f x g x =
∫
f x g x dx( ) ( )0 1
(c)
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Show that L is self-adjoint on D with respect to the inner product of
Recall the integration by parts formula, ∫udv=uv− ∫vdu . Using integration by parts on the integral on the left-hand side of Equation d with u = g(x) and
dv= ⎡⎣⎢d2(αd f dx2 / 2)/dx dx2⎤⎦⎥ gives
Using integration by parts on the remaining integral of Equation e with u = dg/dx and dv= ⎡⎣⎢d(αd f dx2 / 2)/dx dx2⎤⎦⎥ leads to and dg/dx (0) = 0. Thus Equation f reduces to
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d
Using similar steps, it can be shown that for all f(x) and g(x) in D,
f x d
Thus Equation a is proved, and L is self-adjoint on D with respect to the standard inner product.
Example 2.31 The nondimensional partial differential equation for the steady-state temperature distribution, Θ (r) in a bounded three-dimensional region V is
∇ 2 Θ = – f(r) (a)
where r is a position vector from the origin of the coordinate system to a point in the body and ∇ 2 is the Laplacian operator where Bi is the Biot number.
The surface of the region is described by S(r) = 0. The surface of the body is open, and heat transfer occurs though convection, leading to a boundary condition of the form
∇Θ.n = –BiΘ on S (b)
Let Q be the space of all functions defi ned in V. The Laplacian is a linear oper-ator. Defi ne D as the subspace of V consisting of all functions g(r) that satisfy the boundary condition of Equation b. The standard inner product on V is
( ( ), ( ))f r gr =
∫
f( ) ( )r gr dVV
(c)
Show that L = ∇2 is a self-adjoint operator on D with respect to the standard inner product for Q.
Solution If L is self-adjoint, then for any vectors f(r) and g(r) both in D,
(
∇)
=(
∇)
∫
2f gdV∫
f 2g dVV V
(d)
Recall the vector identity,
∇ ∇.
(
g f)
= ∇ ∇ + ∇f. g g 2f (e)9533_C002.indd 127
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Using the identity of Equation e in the integral on the left-hand side of Equa-tion d leads to
( )
∇ =[
−∇ ∇ + ∇ ∇]
∫
V 2f gdV∫
f g g f dVV
. .( ) (f)
The divergence theorem implies
∇ ∇ =
(
∇)
∫
.(g f d) V∫
g f nd. SV S
(g)
where n is a unit r(a) vector normal to S. Equation g leads to
( )
∇ = − ∇ ∇ − ∇∫
2f gdV V∫
f gdV∫
g f ndSV S
. . (h)
Since both f and g are in D, ∇fg . n = − Bif on S and ∇ g · n = −Big on S. Thus Equation h can be written as
( )
∇ = − ∇ ∇ −∫
2f gdV∫
f gdV Bi∫
gfdSV V S
. (i)
In a similar fashion it can be shown that
f
( )
∇ g d = − ∇ ∇f gd − gfd∫
2 V∫
V Bi∫
SV V S
. (j)
The equality of the right-hand sides of Equations i and Equation j proves that L is self-adjoint.
Example 2.32 A Fredholm integral equation is of the form
f x K x y dy g x
x
( ) ( , ) = ( )
∫
0(a)
Equation a can be formulated as Lf = g, where Lf f x K x y dy g x
= ∫x =
0 ( ) ( , ) ( ).
The domain and range of L are C[0,a]. Assuming that the adjoint operator is of the from L*f = ∫x0f x K x y dy( ) *( , ) , determine the form of K*(x,y) using the standard inner product
( , )f g f x g x dx( ) ( )
a
=
∫
0
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Solution If L* is the adjoint of L with respect to the standard inner product, then
f x K x y dy g x dx f
x a
( ) ( , ) ( )
0
0
∫
∫
⎛⎝⎜⎜⎜⎜⎜⎜ ⎞⎠⎟⎟⎟⎟⎟⎟⎟ = (( )x K x y g y dy dx*( , ) ( )x a
0
0
∫
∫
⎛⎝⎜⎜⎜⎜⎜⎜ ⎞⎠⎟⎟⎟⎟⎟⎟⎟ (b)Working with the left-hand side of Equation b by interchanging the order of integration leads to
f y K x y dy g x dx f
x a
( ) ( , ) ( )
0
0
∫
∫
⎛⎝⎜⎜⎜⎜⎜⎜ ⎞⎠⎟⎟⎟⎟⎟⎟⎟ = (( ) ( , ) ( )x K x y g y dxdya y
0
0
∫
∫
(c)Renaming the variables of integration, replacing x by λ and y by τ in Equa-tion c, leads to
(Lf g, ) f( ) ( ) ( , )g K d d
a
=
∫ ∫
τ λ τ λ τ λ τ0 0
(d)
Performing similar steps on the right-hand side of Equation a leads to
( , ) ( ) ( ) ( , )
( ) ( )
* *
f L g f y g x K x y dydx
f g K
x a
=
=
∫
∫
0 0λ τ **( , )τ λ λ τ
τ λ
d d
0
0
∫
∫
(e)
It is clear that Equation d and Equation e are equivalent for all f and g if and only if K x y*( , )=K y x( , ) for all x and y such that 0 ≤ x ≤ aand 0 ≤ y ≤ a.