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Aligning a matrix with its eigen vectors and other questions?

  1. Apr 18, 2012 #1
    Hi,

    I have a square symmetrical matrix A (ugly I know)


    321.1115, -57.5311, -33.9206
    -57.5311, 296.7836, 10.8958
    -33.9206, 10.8958, 382.1050

    which has the eigen values,

    248.8034
    341.6551
    409.5415

    Am I right in saying that A when aligned with its eigen vectors it is,

    248.8034, 0, 0
    0, 341.6551, 0
    0, 0, 409.5415

    ?

    I would also like to transform the matrix so that,

    A11+A22+A33 = 1

    Thanks for any help, I feel like I should know this but have been running around in circles for the past 2 hours.
     
  2. jcsd
  3. Apr 18, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    "aligned with its eigenvectors" is not standard terminology, but I think what you mean is correct. Given any linear transformation, T, from a vector space of dimension n to itself, we can always represent the transformation as a vector space by using a specific ordered basis for the vector space. The idea is that we apply T to each of the basis vectors in turn, writing the result as a linear combination of the basis vectors. For example, if we have a three dimensional vector space with ordered basis [itex]\{v_1, v_2, v_3\}[/itex] then the vector [itex]x_1v_2+ x_2v_2+ x_3v_3[/itex] would be represented by the array
    [tex]\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}[/tex]
    In particular, [itex]v_1= 1v_1+ 0v_2+ 0v_3[/itex] itself is represented by
    [tex]\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}[/tex]

    So if [itex]T(v_1)= a_1v_1+ a_2v_2_+ a_3v_3[/itex] we can write
    [tex]\begin{a_1 & * & * \\ a_2 & * & * \\ a_3 & * & * \end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a_1 \\ a_2\\ a_3\end{bmatrix}[/tex]
    where the "*" in the second and third columns can be anything.

    That is, the result result of T applied to basis vector [itex]v_i[/itex] gives the ith column of the matrix representation. In particular, if the basis vectors are eigenvalues of T, then [itex]Tv_1= \lamba_1 v_1+ 0 v_2+ 0v_3[/itex], [itex]Tv_2= 0v_1+ \lambda_2v_2+ 0v_3[/itex], and [itex]Tv_3= 0v_1+ 0v_2+ \lambda_3v_3[/itex] so the matrix representation, in that basis, is
    [tex]\begin{bmatrix}\lambda_2 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3\end{bmatrix}[/tex]

    To reduce those diagonal elements to 1 is to divide each eigenvector by the corresponding eigenvalue. If [itex]v_1[/itex] is an eigenvector of T with eigenvalue [itex]\lambda_1[/itex] then
    [tex]T\frac{v_1}{\lambda_1}= \frac{1}{\lambda_1}Tv_1= \frac{1}{\lambda_1}\lambda_1 v_1= v_1[/tex]
    so representing T as as matrix by using basis vectors [itex]v_1/\lambda_1[/itex], [itex]v_2/\lambda_2[/itex], and [itex]v_3/\lambda_3[/itex] will give the identity matrix.
     
  4. Apr 19, 2012 #3
    Thanks for the reply.

    Once the matrix (A) has been transformed as above how can I scale it such that,

    trace(A)=1

    ?
     
    Last edited: Apr 19, 2012
  5. Apr 19, 2012 #4
    OK panic over I just divide the aligned matrix by its trace.

    Thanks for your time.
     
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