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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⎦

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