Lef A be a real nxm UE V! uhere madsix and lel At= vE'uT, where booh unitary ,E is U is mam, V cs nan, A = mxn and E's nxm have ohe borm an's -1 Σ ro and Σ 232.00 [xJ くだい Wi'th Gと52 … show: * = A b a least squares solution is to Ax = b Previously we used * = CA') b bor our least square solution, that columns of the A' b. Bor our %3D we had the restriction data" matrix A were but cndepen dend NSCA) guarantees thad NS CATA) ={0} this %3D to see that A* =(nTA ) linear indepen dent columns . It is not hard has くX
Lef A be a real nxm UE V! uhere madsix and lel At= vE'uT, where booh unitary ,E is U is mam, V cs nan, A = mxn and E's nxm have ohe borm an's -1 Σ ro and Σ 232.00 [xJ くだい Wi'th Gと52 … show: * = A b a least squares solution is to Ax = b Previously we used * = CA') b bor our least square solution, that columns of the A' b. Bor our %3D we had the restriction data" matrix A were but cndepen dend NSCA) guarantees thad NS CATA) ={0} this %3D to see that A* =(nTA ) linear indepen dent columns . It is not hard has くX
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.9: Properties Of Determinants
Problem 46E
Related questions
Question
![Lef A be a real nxm
UE V! uhere
madsix and lel At= vE'uT, where
booh unitary ,E is
U is mam, V cs nan,
A =
mxn and E's
nxm have ohe borm
an's
-1
Σ
ro and
Σ
232.00
[xJ
くだい
Wi'th
Gと52 …
show:
* = A b
a least squares solution
is
to
Ax = b
Previously we used * = CA') b bor our
least square solution,
that columns of the
A' b. Bor our
%3D
we had the restriction
data" matrix A were
but
cndepen dend
NSCA)
guarantees thad
NS CATA) ={0}
this
%3D
to see that A* =(nTA
)
linear indepen dent columns .
It is not hard
has
くX](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fdbb1a7-d573-479c-8d2a-d1c6b6fb9ae3%2F49a4e04a-66dd-4c70-bc3f-df9ea6dff1ab%2Fu4mldh9_processed.png&w=3840&q=75)
Transcribed Image Text:Lef A be a real nxm
UE V! uhere
madsix and lel At= vE'uT, where
booh unitary ,E is
U is mam, V cs nan,
A =
mxn and E's
nxm have ohe borm
an's
-1
Σ
ro and
Σ
232.00
[xJ
くだい
Wi'th
Gと52 …
show:
* = A b
a least squares solution
is
to
Ax = b
Previously we used * = CA') b bor our
least square solution,
that columns of the
A' b. Bor our
%3D
we had the restriction
data" matrix A were
but
cndepen dend
NSCA)
guarantees thad
NS CATA) ={0}
this
%3D
to see that A* =(nTA
)
linear indepen dent columns .
It is not hard
has
くX
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