Linear Algebra: show it is diagonalizable and find eigenbasis

[braindead101]

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?

 

[HallsofIvy]

Apply the definition of "eigenvector": any eigenvector, v, of A, corresponding to eigenvalue \lambda satisfies Av= \lambda v.
For eigenvector \lambda= \pi
\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)
Which gives the equations \pi x= \pi x, 2z= \pi y, and -2y= \pi z. The first is true for any x, the last two are only true if y= z= 0. Any eigenvector corresponding to \lambda= 1 is of the form (x, 0, 0) and the eigenspace is spanned by (1, 0, 0).

For eigenvalue 2i,
\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)
which gives us equations \pi x= 2i x, 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 \lambda= -2i to you.