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braindead101

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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?

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?

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