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    KathMae Boa
    Eigenvalues and eigenvectors are values and vectors that satisfy the equation Ax = λx, where A is a matrix, x is a vector, and λ is a scalar value. To find the eigenvalues, we solve the characteristic equation det(A - λI) = 0. The eigenvectors corresponding to each eigenvalue are the non-trivial solutions to the equation (A - λI)x = 0. We can find the eigenvectors by reducing (A - λI) to row-echelon form. This process provides the eigenvalues and eigenvectors of a given matrix.

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    Eigenvalues and eigenvectors are values and vectors that satisfy the equation Ax = λx, where A is a matrix, x is a vector, and λ is a scalar value. To find the eigenvalues, we solve the characteristic equation det(A - λI) = 0. The eigenvectors corresponding to each eigenvalue are the non-trivial solutions to the equation (A - λI)x = 0. We can find the eigenvectors by reducing (A - λI) to row-echelon form. This process provides the eigenvalues and eigenvectors of a given matrix.

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    Eigenvalues and eigenvectors are values and vectors that satisfy the equation Ax = λx, where A is a matrix, x is a vector, and λ is a scalar value. To find the eigenvalues, we solve the characteristic equation det(A - λI) = 0. The eigenvectors corresponding to each eigenvalue are the non-trivial solutions to the equation (A - λI)x = 0. We can find the eigenvectors by reducing (A - λI) to row-echelon form. This process provides the eigenvalues and eigenvectors of a given matrix.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
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    98 views11 pages

    9 PDF

    Uploaded by

    KathMae Boa
    Eigenvalues and eigenvectors are values and vectors that satisfy the equation Ax = λx, where A is a matrix, x is a vector, and λ is a scalar value. To find the eigenvalues, we solve the characteristic equation det(A - λI) = 0. The eigenvectors corresponding to each eigenvalue are the non-trivial solutions to the equation (A - λI)x = 0. We can find the eigenvectors by reducing (A - λI) to row-echelon form. This process provides the eigenvalues and eigenvectors of a given matrix.

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    Eigenvalues

    Eigenvalues and Eigenvectors


     Let A be an nxn matrix and consider the vector
    equation:
    Ax = x
     A value of  for which this equation has a solution
    x≠0 is called an eigenvalue of the matrix A.
     The corresponding solutions x are called the
    eigenvectors of the matrix A.
     where  is a scalar.
    Solving for eigenvalues
    Ax=x
    Ax - x = 0
    (A- I)x = 0
     This is a homogeneous linear system, homogeneous
    meaning that the RHS are all zeros.
     For such a system, a theorem states that a solution
    exists given that det(A- I)=0.
     The eigenvalues are found by solving the above
    equation.
    Eigenvalue and Eigenvector)
     The primary problem of interest for linear systems of
    the form is to determine those values of for which

    the system has a nontrivial solution ; such a value of

    is called a characteristic value or an eigenvalue of A.

     If  is an eigenvalue of A, then the nontrivial solutions


    are called the eigenvectors of A corresponding to .

    Example
    1
     The vector x= is an eigenvector of
    2
    A= 3 0
    8 -1

    3 0 1 3
    Since Ax = 8 –1 2 = 6 =3x.

    The eigenvalue =3.


    How to find :
     Recall that Ax=x. Then (A-I)x=0.
     Eigenvalues can be found be solving the
    characteristic equation det(A-I)=0.

    3 0 3- 0
     Example : A= 8 -1 A-I = 8 -1-

    3- 0
    8 -1-
     Det( ) =(3-)(-1-)=0
    =3 or =-1
    Finding eigenvectors associated with
    eigenvalues.
     Using the example =3 or =-1. There is an eigenvector
    associated with each eigenvalue.

     Let’s find the eigenvector x= x1 associated with 3.


    x2
     Recall that Ax=x. Then Ax=3x or A-3I=0.
    3 0 0 0
     A= 8 -1 A-3I = 8 -4
     Solve A-3I=0. We get that x2 is free and that x1=.5x2.
    Thus the solution is .5
    .5x2
     The eigenvector is 1 x2
    Finding eigenvectors associated with
    eigenvalues.


    x
    Let’s find the eigenvector x= 1 associated with -1.
    x2

     Then Ax=-x or A+I=0.


    4 0
    3 0
    8 -1 80
     A= A+I =
     Solve A+I=0. We get that x2 is free and that x1=0. Thus
    0
    the solution is 0 and the eigenvector is
    x2 1
    Example 2: Find Eigenvalues and Eigenvectors

    1 2 1
    A= 6 -1 0
    -1 -2-1

    Solution:
    To expand the determinant in the characteristic equation.

    1-  2 1
    det (A-  I)= 6 -1-  0 =0
    -1 -2 -1- 
    - -
    3 2+12=0 or (+4)(-3)=0. Hence the
    eigenvalues are: 1=o, 2=-4, 3=3.

    To find the eigenvectors we must reduce(A-I 0), three times


    corresponding to three eigenvalues.

    For 1=0 , we have


    1- 0 2 1 -6R1+R2 1 2 1
    R1 +R3 -6
    det (A- 0 I)= 6 -1- 0 0 0 -13
    -1 -2 -1- 0 0 0 0
    -1/13R2 1 2 1 -2R2 +R1 1 0 1/13
    0 1 6/13 0 1 6/13
    0 0 0 0 0 0

    Thus, we see that k1=-1/13k3 and k2=-6/13k3.


    Choosing k3=- 13 gives the eigenvector,
    1
    K1= 6
    -13

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