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Eigenvalues and characteristic polynomials

  1. Oct 26, 2012 #1
    Hello guise.

    I am familiar to a method of diagonalizing an nxn-matrix which fulfills the following condition:
    the sum of the dimensions of the eigenspaces is equal to n.

    As to the algorithm itself, it says:

    1. Find the characteristic polynomial.
    2. Find the roots of the characteristic polynomial.
    3. Let the eigenvectors [itex]v_{i}[/itex] be the column vectors of some matrix S.
    4. Let the eigenvalues [itex]\lambda_{i}[/itex] be the elements of some diagonal matrix, ordered to CORRESPOND the order of the eigenvectors in S.
    5. Our Diagonalization of A should be:
    [itex]A = S \cdot A \cdot S^{-1} = (v_{1}...v_ {i}...v_ {n}) \cdot (\lambda_{1}...\lambda_{i}...\lambda_{n}) \cdot S^{-1}[/itex]

    My question is: how do I find at least one such matrix A Corresponding to some randomly created polynomial of degree [itex]m[/itex] with integer roots? If it is too difficult to solve this for an arbitrary [itex] m [/itex], that's okay. But let's say for [itex] m = 5[/itex]? Or for the much simpler case of [itex] m = 2[/itex]?
    Is this somehow related to the quadratic form?
     
  2. jcsd
  3. Oct 26, 2012 #2

    HallsofIvy

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    Given the m roots, construct the m by m diagonal matrix, D, having those roots on the main diagonal. For any invertible matrix, P, [itex]A= PDP^{-1}[/itex] will have characteristic polynomial with those roots.
     
  4. Oct 26, 2012 #3
    Jesus christ, I feel stupid. Thanks.
     
  5. Oct 26, 2012 #4

    mathwonk

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    thats how people make up exercises and test problems.
     
  6. Oct 27, 2012 #5
    A follow-up question: how do I find a matrix with integer elements whose inverse also has integer elements? ^^
     
  7. Oct 27, 2012 #6

    Hurkyl

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    You need its determinant to be 1 or -1.

    Any integer matrix is a product of integer elementary matrices corresponding to the row operations:
    • Adding an integer multiple of one row to another
    • Swapping two rows
    • Multiplying a row by a constant

    As long as you only use 1 or -1 for that last one, all three sorts are invertible.
     
  8. Oct 27, 2012 #7
    Hahahaha! How brilliant. Okay. Now I can finally randomly create integer matrices A such that [itex]A = SDS^{-1}[/itex] are integer matrices as well! Other questions:

    -How do I find a matrix A with integer elements, which has an inverse with rational elements? I get the feeling that it always is that a "rational" matrix always has a "rational" inverse [but I do not know how to show it]?

    -How do I find a matrix A with rational elements which has an inverse with integer elements?

    -How do I prove that a real matrix never has an inverse with complex elements?
     
    Last edited by a moderator: Oct 27, 2012
  9. Oct 27, 2012 #8

    HallsofIvy

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    You can't create a matrix A such that this will work for all S. You need both A and S having determinants 1 or -1.

    The inverse of matrix A is the matrix of cofactors divided by the determinant of A so, yes, since the rational numbers is closed under addition, subtraction, multiplication, and division by non-zero numbers, the inverse of a matrix with rational entries (including integers) will always have rational entries.

    An obvious way is to start with an invertible matrix with integer coefficients and let A be its inverse! More generally if the determinant of A is a common multiple of its denominators (and so an integer) then its inverse will have integer entries.

    Again, every element of A-1 is its cofactor divided by the matrix of A. Since the real numbers is closed under the four basic operations, that will always give real numbers.
     
  10. Oct 27, 2012 #9


    If you don't have the eigenvalues you use the companion matrix of a monic polynomial:
    $${}$$
    $$\left(\begin{array} {}0&1&0&0&...&0\\0&0&1&0&...&0\\...&...&...&...&...&...\\0&0&0&...&0&1\\\!\!-c_0&\!\!-c_1&\!\!-c_2&\!\!-c_3&...&\!\!-c_{n-1}\end{array}\right)$$

    The coolest feature of the above matrix is that [itex]\,f(x)=c_0+c_1x+...+c_{n-1}x^{n-1}+x^n\,[/itex] is not only its

    characteristic but also its minimal polynomial...!

    DonAntonio
     
  11. Oct 27, 2012 #10
    Very interesting! But I am somewhat confused. How do I actually use it? I was wondering how to find my matrix A out of some given polynomial - but that would require me to know it's roots, due to the design of the algorithm. For instance, if I have my matrix {{2,1},{1,2}} then it's characteristic polynomial is x^2-4x+3. If I use the formula you gave me, it's corresponding matrix would be {{0,1},{0,0},{-3,-2}}, which is non-square and therefore not a matrix A which fulfills the criteria of the algorithm? Or am I completely mistaken?
     
    Last edited: Oct 27, 2012
  12. Nov 4, 2012 #11
    [itex] \left( \begin{array}{cc}
    0 & 1\\ -3 & 4\\ \end{array} \right) ?
    [/itex]
     
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