Let F=R₁₂ C three Given conditions for a (a) S (b) s is b = orth ís 19 S 15 a any LS solution A = M (F) vector solution of of Oh pros of b an (exact) solution and bef", the following sef" are equivalent: the system Ax=b the system Ax=b', where cal(A) & col(A) = R(T) of the system A*Ax= Ab
Let F=R₁₂ C three Given conditions for a (a) S (b) s is b = orth ís 19 S 15 a any LS solution A = M (F) vector solution of of Oh pros of b an (exact) solution and bef", the following sef" are equivalent: the system Ax=b the system Ax=b', where cal(A) & col(A) = R(T) of the system A*Ax= Ab
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Main Thm
Let F=R ov (
Given
three conditions for a
on LS
(a) s
ibi s is
b' = orth
1C) S is
is a
solutions
any
a
A = Mmx (F)
LS
solution
pros
of b
an (exact) solution of
vector
solution of
of
Oh
SEF"
and be FM, the following
equivalent:
the
are
the
system Ax=b
the system Ax=b', where
cal(A) & col(A) = R(T)
system A*Ax=A* b
Transcribed Image Text:3. Let F R or C, WC F be an m-dimensional subspace, and Pw: FnF be the
orthogonal projection onto W. Choose a basis {w₁,..., wm} for W, and let A be the
nxm matrix whose ith column is w; for i = 1, ..., m. Let v € Fn. Since Pw(v) E W
and {w₁,..., wm} is a basis for W, we can write Pw (v) = a₁w₁ + + amwm for
some unique a₁,..., am € F.
(a) Use the main theorem on least squares solutions to show that
the set of all least squares solutions to Ax = v
= Pw (v)
A*v.
=
the set of all solutions to Ax=
= the set of all solutions to A*Ax =
(b) Use the first equality in (a) to show that the system Ax = v has a unique least
a1
·0
Am
squares solution, which is
(c) Now use the second equality in (a) and the result of (b) to show that the
matrix A* A is invertible, and
(A*A)-¹A* v=
(0)
Am
(d) (Formula for the orthogonal projection matrix). Use the result of (c)
to show that the matrix representation of Pw relative to the standard basis
for Fn is A(A*A)−¹A*.
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