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Math Help - The arithmetic and geometric multiplicity of an eigenvector

  1. #1
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    The arithmetic and geometric multiplicity of an eigenvector

    I have a matrix and I am supposed to calculate the eigenvalue, eigenvector and then give arithmetic and algebraic multiplicity.I can calculate the eigenvalues and the eigenvectors but I don't know how to calculate the arithmetic and geometric multiplicity. Tried to google it but it was a bit ambiguous for my understanding. The matrix is |2 1| and eigenvalues are 1, 4 and eigenvectors are (-1 1) and (1 2) . |2 3|
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  2. #2
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    Quote Originally Posted by Keep View Post
    I have a matrix and I am supposed to calculate the eigenvalue, eigenvector and then give arithmetic and algebraic multiplicity.I can calculate the eigenvalues and the eigenvectors but I don't know how to calculate the arithmetic and geometric multiplicity. Tried to google it but it was a bit ambiguous for my understanding. The matrix is |2 1| and eigenvalues are 1, 4 and eigenvectors are (-1 1) and (1 2) . |2 3|
    So your characteristic equation is (x- 1)(x- 4)= 0? The "algebraic multiplicity" of an eigenvalue, \lambda is the multiplicity of the factor (x- \lambda) in the characteristic polynomial. Here that is 1 for both eigenvalues. The geometric multiplicity is the number of independent eigenvectors corresponding to that eigenvalue and can be from 1 to the algebraic multiplicity of the eigenvalue. Here, of course, that is 1.
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  3. #3
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    Hi

    The characteristic polynomial is \lambda - 4)" alt="\lambda^2 - 5 \lambda + 4 = (\lambda-1)\\lambda - 4)" />

    The algebraic multiplicity of each eigenvalue e is the exponent of (\lambda - e) ; here it is 1 for each of the 2 eigenvalues
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  4. #4
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    I think I got the algebraic multiplicity. It is basically how many equal eigenvalues are there. So if we have in a 3X3 matrix (x-4)^3=0 then the algebraic multiplicity is 3 because we have 3 eigenvalues of the same value? But I still don't get the geometric one. I would like to go by the definition of how many independent eigenvectors corresponding to the eigenvalues are there since I know how to find out the linear independence of vectors (using determinant). But we usually work out only one eigenvector, which is a column vector for a matrix, and to calculate the linear independence of vectors, we use not just one column but a family of them so how can you calculate the linear independence of one single column vector? And what does corresponding mean in this sense? As an example, I have (2 0 0; 1 3 0; -1 2 5) matrix. The eigenvalues of these 3x3 matrix are 2,3,5 and the algebraic multiplicity is therefore 1 for each eigenvalue; the eigenvector for the eigenvalue 2 is (1, -1, 1), for eigenvalue 3 is (0, -1, 1) and for eigenvalue 5 is (0, 0, 1). But how do you calculate the geometric multiplicity then?
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  5. #5
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     <br />
\begin{pmatrix}2 & 0 & 0\\1 & 3 & 0\\-1 & 2 & 5\end{pmatrix}<br />

    The characteristic polynomial is (x-2)(x-3)(x-5).
    The algebraic multiplicity of eigenvalue 2 is 1.
    The algebraic multiplicity of eigenvalue 3 is 1.
    The algebraic multiplicity of eigenvalue 5 is 1.

    To find the eigenvector corresponding to eigenvalue 2 you need to solve for x, y, z (not all equal to 0)

     <br />
 \begin{pmatrix}0 & 0 & 0\\1 & 1 & 0\\-1 & 2 & 3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}<br />

     <br />
 \begin{pmatrix}0 \\x+y\\-x + 2y + 3z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}<br />

    x+y = 0<br />
    -x + 2y + 3z = 0<br />

    Let's parametrize x = a \implies y = -a \implies z = a which leads to eigenvector
     <br />
 \begin{pmatrix}a \\-a\\a\end{pmatrix} = a\begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix}<br />

    The eigenvectors are characterized by one direction \begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix}<br />
  .
    All the eigenvectors are parallel.
    The geometric multiplicity is 1.


    Now let's modify a little bit the previous example
     <br />
\begin{pmatrix}2 & 0 & 0\\1 & 2 & 0\\-1 & 2 & 5\end{pmatrix}<br />

    The characteristic polynomial is (x-2)^2(x-5).
    The algebraic multiplicity of eigenvalue 2 is 2.
    The algebraic multiplicity of eigenvalue 5 is 1.

    To find the eigenvector corresponding to eigenvalue 2 you need to solve for x, y, z (not all equal to 0).

     <br />
 \begin{pmatrix}0 & 0 & 0\\1 & 0 & 0\\-1 & 2 & 3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}<br />

     <br />
 \begin{pmatrix}0 \\x\\-x + 2y + 3z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}<br />

    x = 0<br />
    -x + 2y + 3z = 0<br />

    x = 0<br />
    2y + 3z = 0<br />

    Let's parametrize y = 3a \implies z = -2a which leads to eigenvector
     <br />
  \begin{pmatrix}0 \\3a\\-2a\end{pmatrix} = a\begin{pmatrix}0 \\ 3 \\ -2\end{pmatrix}<br />

    The eigenvectors are characterized by one direction \begin{pmatrix}0 \\ 3 \\ -2\end{pmatrix}<br />
   .
    All the eigenvectors are parallel.
    The geometric multiplicity is 1.

    You could find examples where the algebraic multiplicity and the geometric multiplicity are equal to 2.
    In any case 1 \leq geometric multiplicity \leq algebraic multiplicity
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  6. #6
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    Quote Originally Posted by running-gag View Post
     <br />
\begin{pmatrix}2 & 0 & 0\\1 & 3 & 0\\-1 & 2 & 5\end{pmatrix}<br />

    The characteristic polynomial is (x-2)(x-3)(x-5).
    The algebraic multiplicity of eigenvalue 2 is 1.
    The algebraic multiplicity of eigenvalue 3 is 1.
    The algebraic multiplicity of eigenvalue 5 is 1.

    To find the eigenvector corresponding to eigenvalue 2 you need to solve for x, y, z (not all equal to 0)

     <br />
 \begin{pmatrix}0 & 0 & 0\\1 & 1 & 0\\-1 & 2 & 3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}<br />

     <br />
 \begin{pmatrix}0 \\x+y\\-x + 2y + 3z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}<br />

    x+y = 0<br />
    -x + 2y + 3z = 0<br />

    Let's parametrize x = a \implies y = -a \implies z = a which leads to eigenvector
     <br />
 \begin{pmatrix}a \\-a\\a\end{pmatrix} = a\begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix}<br />

    The eigenvectors are characterized by one direction \begin{pmatrix}1 \\ -1 \\ 1\end{pmatrix}<br />
  .
    All the eigenvectors are parallel.
    Up to here I am with you. But we part ways a little bit from here.
    Quote Originally Posted by running-gag View Post
    The geometric multiplicity is 1.
    My question is how do we know it is 1? I understand how you can get the algebraic multiplicity but how do you work out the geometric one? If we have everything; the characteristic polynomial, the eigenvalue, the eigenvector, the algebraic multiplicity. Now how do you get the geometric multiplicity?


    Quote Originally Posted by running-gag View Post
    Now let's modify a little bit the previous example
     <br />
\begin{pmatrix}2 & 0 & 0\\1 & 2 & 0\\-1 & 2 & 5\end{pmatrix}<br />

    The characteristic polynomial is (x-2)^2(x-5).
    The algebraic multiplicity of eigenvalue 2 is 2.
    The algebraic multiplicity of eigenvalue 5 is 1.

    To find the eigenvector corresponding to eigenvalue 2 you need to solve for x, y, z (not all equal to 0).

     <br />
 \begin{pmatrix}0 & 0 & 0\\1 & 0 & 0\\-1 & 2 & 3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}<br />

     <br />
 \begin{pmatrix}0 \\x\\-x + 2y + 3z\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}<br />

    x = 0<br />
    -x + 2y + 3z = 0<br />

    x = 0<br />
    2y + 3z = 0<br />

    Let's parametrize y = 3a \implies z = -2a which leads to eigenvector
     <br />
  \begin{pmatrix}0 \\3a\\-2a\end{pmatrix} = a\begin{pmatrix}0 \\ 3 \\ -2\end{pmatrix}<br />
    All fine and well understood but can I parametrise it this way? We have 2y+3z=0. Can I say z=a and then go ahead and solve 2y= -3a. And then I will get y=-1.5a. To remove the decimal, multiply each by 2 so we have y= -3a and z=2a so we will have \begin{pmatrix}0 \\ -3 \\ 2\end{pmatrix}. Is this also valid or?

    Quote Originally Posted by running-gag View Post
    The eigenvectors are characterized by one direction \begin{pmatrix}0 \\ 3 \\ -2\end{pmatrix}<br />
   .
    All the eigenvectors are parallel.
    The geometric multiplicity is 1.
    Again, I would be happy to know how I can work out the geometric multiplicity. I have understood everything else except how I can get the geometric multiplicity. Also you say all the eigenvectors are parallel? We have only one eigenvector namely \begin{pmatrix}0 \\ 3 \\ -2\end{pmatrix}<br />
   so why do you say all the eigenvectors?

    Quote Originally Posted by running-gag View Post
    You could find examples where the algebraic multiplicity and the geometric multiplicity are equal to 2.
    In any case 1 \leq geometric multiplicity \leq algebraic multiplicity
    Thank you. You really explained well and helped me understand better except for the geometric multiplicity.
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  7. #7
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    Hello, i am new here, and this thread comes up when you type in geometric multiplicity into google. A firm answer was never given as too how the geometric multiplicity is calculated.
    I know this is an old thread, but could someone please come out with a firm answer in plain english? there is nothing out there that gives it in an easy to understand way.
    Thank You
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