Linear Transformation Examples Matrix Eigenvalue problems
Linear Transformation Examples Matrix Eigenvalue problems
Linear Transformation Examples Matrix Eigenvalue problems
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Md55<br />
<strong>Eigenvalue</strong>s and Eigenvectors<br />
● Terminology:<br />
Ax = λx matrix eigenvalue problem<br />
A value of λλ for which x (≠ 0) is a solution eigenvalue,<br />
(also known as characteristic value)<br />
Solutions x (≠ 0) corresponding to a λ called eigenvectors<br />
The set of eigenvectors is called the spectrum of A<br />
Largest of absolute values of eigenvalues is spectral radius of A<br />
● Determination of eigenvalues and eigenvectors:<br />
Ax = λx = λIx, where I is the identity matrix<br />
(A - λI)x = 0, homogeneous linear system with non-trivial solution<br />
(x ≠ 0) if and only if D(λλ) = det(A - λI) = 0<br />
Md56<br />
Example<br />
Ax = x A =<br />
x<br />
x<br />
x<br />
x x x<br />
x x x<br />
− ⎡<br />
⎢<br />
⎣<br />
⎤<br />
− ⎥ =<br />
⎦<br />
⎡<br />
● Determination of eigenvalues :<br />
5<br />
λ ,<br />
2<br />
2<br />
1 ⎤<br />
, ⎢ ⎥,<br />
so that<br />
2 ⎣ 2 ⎦<br />
− 5 1 + 2 2 = λ 1<br />
2 1 − 2 2 = λ 2<br />
− − x + x =<br />
A− I x =<br />
x + − − x =<br />
D = A− I = −<br />
( 5 λ)<br />
1 2 2 0<br />
( λ ) 0,<br />
2 1 ( 2 λ)<br />
2 0<br />
and<br />
5−λ ( λ) det( λ )<br />
2<br />
2<br />
= 0<br />
−2−λ Characteristic polynomial<br />
2<br />
D(<br />
λ) = ( −5−λ)( −2−λ) − 4 = λ + 7λ + 6= 0<br />
Whence, ( λ + 1)( λ + 6) = 0, so that λ = −1, − 6; i.e. λ =− 1, λ =−6.<br />
● Determination of an eigenvector:<br />
1 2<br />
For<br />
− + =<br />
λ = λ = − :<br />
− =<br />
and these equations are linearly dependent<br />
with a solution : = . This determines an eigenvector corresponding to λ = −<br />
up to a scalar multiple. If we choose x 1 = , we obtain eigenvector x = e1<br />
= .<br />
⎡ ⎤ ⎡<br />
⎢ ⎥ = ⎢<br />
⎣ ⎦ ⎣<br />
⎤<br />
4x1 2x2 0<br />
1 1<br />
2x1 x2<br />
0<br />
x2 2x1 1 1<br />
x1<br />
1<br />
1<br />
x ⎥<br />
2 2⎦