Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Eigenvalue problem, eigenvector, determinant, hermitian matrices, eigen decomposition, rayleigh quotient scaling
Typology: Study notes
1 / 35
This page cannot be seen from the preview
Don't miss anything!
-^ Suppose we if eliminate the components of
x
from A
x =
0 using Gaussian Elimination without
pivoting. • We do the kth forward elimination step bysubtracting a
times row k from ajk
times row j,kk^
for j=k,k+1,…n. • Then at the end of forward elimination we haveU x
=^0
, and U
nn^
is the determinant of A, det(A).
-^ For nonzero solutions to A
x =
0 we must have
det(A) = 0. • Determinants are defined only for squarematrices.
-^ Suppose we have a^3
μ3 matrix.
⎞⎟ ⎟ ⎟⎠
⎛⎜ =⎜ ⎜⎝
33 32 31
23 22 21
13 12 11
a a a
a a a
a a a A
-^ So A
x =
0 is the same as: a^11
x^ +a^1
x 12
+a 2
x 13
= 0 3
a^21
x^ +a^1
x 22
+a 2
x 23
= 0 3
a^31
x^ +a^1
x 32
+a 2
x 33
= 0 3
-^ Step k=2:
) times equation 3. and so:
-^ Minor M
of matrix A is the determinant of theij^
matrix obtained by removing row i and column jfrom A. • Cofactor C
= (-1)ij^
i+j M
ij
-^ If A is a 1
μ1 matrix then det(A)=a
. 11 -^ In general,
ij n j
where i can be any value i=1,…n.
⎞ ⎟ ⎟ ⎠ ⎛ ⎜ = ⎜ ⎝
2 1
3 4
⎞ ⎟⎟ ⎠
⎛ ⎜⎜ ⎝
− − = −
λ λ
λ^
2 1
3
4 I A
0 ) 3 )( (^1) ( ) (^2) )( (^4) (
0 )
det(
= − − − ⇒ = −
A
2
Eigenvalues of Positive Definite
Matrices
-^ A square matrix for which A = A
H^ is said to be an
Hermitian
matrix.
-^ If A is real and Hermitian it is said to be^ symmetric
, and A = A
T^.
-^ Every Hermitian matrix is positive definite. •^ Every eigenvalue of an Hermitian matrix is real. •^ Different eigenvectors of an Hermitian matrix areorthogonal to each other, i.e., their scalarproduct is zero.
-^ Let
l^1
,l^2
,…,
ln^
be the eigenvalues of the n
μn
matrix A and
x^1
, x^2
,…,
x n^
the corresponding
eigenvectors. • Let
L be the diagonal matrix with
l^1
,l^2
,…,
ln^
on
the main diagonal. • Let X be the n
μn matrix whose jth column is
x .j
-^ Then AX = X
L, and so we have the
eigen
decomposition
of A:^ A = X
LX
0
-^ We know that A
k^ = X
kL
-1X , so:
y k^
= A
k y^0
= X
kL -1X y^0
-^ Now we have:
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞⎟ ⎟^ =⎟ ⎟ ⎟ ⎠
⎛ ⎜ ⎜ =⎜ ⎜ ⎜ ⎝ Λ
k n^ k
k k k k n
k k k
1
2 1 1
2 1
1
λ^ λ
λ λ λ λ
λ λ
O
O
-^ The terms on the diagonal get smaller inabsolute value as k increases, since
l^1
is the
dominant eigenvalue.
