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Linear Algebra: show it is diagonalizable and find eigenbasis

  1. Feb 14, 2008 #1
    Let T1 be the C-vector space with basis B = (1, cosx, sinx). Define J: T1->T1 by (Jf)(x) = integ(0->pi) f(x-t)dt. Show that J is diagonalizable and find an eigenbasis.

    J(1) = integ(0->pi) 1 dt
    J(1) = t | (0->pi)
    J(1) = pi

    J(cos(x)) = integ(0->pi) cos(x-t) dt
    J(cos(x)) = - sin (x-t) | (0->pi)
    J(cos(x)) = - sin (x-pi) + sin(x)
    J(cos(x)) = - (sin(x) cos(pi) - cos(x) sin (pi)) + sin(x)
    J(cos(x)) = - (-sin(x)) + sin(x)
    J(cos(x)) = 2sin(x)

    J(sin(x)) = integ(0->pi) sin(x-t) dt
    J(sin(x)) = cos (x-t) | (0->pi)
    J(sin(x)) = cos (x-pi) - cos (x)
    J(sin(x)) = cos(x) cos (pi) + sin(x) sin(pi) - cos (x)
    J(sin(x)) = -2cos(x)

    [J]T1 = pi 0 0
    0 0 -2
    0 2 0

    v = lambda
    So, det [J-vI] = (pi-v)(v^2 +4)
    So, pi - v = 0 or v^2 +4 = 0
    v = pi, v = 2i v = -2i

    how do i go about looking for eigenbasis?
     
    Last edited: Feb 14, 2008
  2. jcsd
  3. Feb 14, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Apply the definition of "eigenvector": any eigenvector, v, of A, corresponding to eigenvalue [itex]\lambda[/itex] satisfies [itex]Av= \lambda v[/itex].
    For eigenvector [itex]\lambda= \pi[/itex]
    [tex]\left(\begin{array}{ccc} \pi & 0 & 0 \\ 0 & 0 & 2 \\ 0 & -2 & 0\end{array}\right)\left(\begin{array}{c} x \\ y \\ z \end{array}\right)= \left(\begin{array}{c} \pi x \\ \pi y \\ \pi z\end{array}\right)[/tex]
    Which gives the equations [itex]\pi x= \pi x[/itex], [itex]2z= \pi y[/itex], and [itex]-2y= \pi z[/itex]. The first is true for any x, the last two are only true if y= z= 0. Any eigenvector corresponding to [itex]\lambda= 1[/itex] is of the form (x, 0, 0) and the eigenspace is spanned by (1, 0, 0).

    For eigenvalue 2i,
    [tex]\left(\begin{array}{ccc} \pi & 0 & 0 \\ 0 & 0 & 2 \\ 0 & -2 & 0\end{array}\right)\left(\begin{array}{c} x \\ y \\ z \end{array}\right)= \left(\begin{array}{c} 2i x \\ 2i y \\ 2i z\end{array}\right)[/tex]
    which gives us equations [itex]\pi x= 2i x[/itex], 2z= 2i y, and -2y= 2i z. The first equation is true only for x= 0 and the last two are true as long as z= i y. The eigenspace is any vector of the form (0, y, iy) and is spanned by (0, 1, i).

    I'll leave [itex]\lambda= -2i[/itex] to you.
     
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