Formula Sheet For Linear Algebra - Final
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1
2Formula Sheet for Linear Algebra Final
6.3 Orthonormal bases; Gram-Schmidt process; QR-decomposition
Apply the Gram-Schmidt process to a set
,
, . . . ,
of vectors to obtain an orthonor-
1
2
mal set
,
, . . . ,
of vectors with the same span as the original set. If the original set
1
2
was a basis, then the final set will be an orthonormal basis. To find the vector
, subtract
from
its projections onto the vectors
, . . . ,
(to get
) and then normalize this
1
1
vector to get
:
=
,
,
,
,
1
1
2
2
1
1
=
.
Use the results of Gram-Schmidt orthogonalization to write a matrix A with linearly
independent columns as a product
A = QR,
where Q has orthonormal columns and R is an invertible upper triangular matrix. With
A = [
],
1
2
we set
Q = [
]
1
2
and
,
,
,
1
2
1
3
1
1
0
,
,
2
3
2
2
0
0
,
R =
.
3
3
. .
. .
. .
. .
.
.
.
.
0
0
0
Note that
=
,
=
,
.
6.4 Least squares approximation
If W is a finite-dimensional subspace of an inner product space V , and if
is a vector in
V , then the vector in W closest to
is proj
.
Given a matrix A
M
, the set of all vectors of the form A is the span of the column
vectors
, . . . ,
of A. Thus A =
has a solution
if and only if
1
span
, . . . ,
.
1
If
/ span
, . . . ,
, then A
=
does not have any solutions
. However, the
1
associated normal system
A A = A
always has a solution . A solution
of the normal system satisfies proj
= A , so
that A is as close to
as possible. If
, . . . ,
are linearly independent, then A A
1
will be invertible, so we have
1
= (A A)
A
.
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