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TranscendArcu is offline
Feb26-12, 02:24 PM
P: 288
Quote Quote by Dick View Post
You are computing the characteristic polynomial. Once you find that the roots are the eigenvalues. [itex](1-t)[(1-t)(2-t)-2[/itex] is right, the other side isn't. You want to factor it.
Should I write: [itex](1-t)[(1-t)(2-t)-2 = -(t-3)(t-1)(t)[/itex]. This is the characteristic polynomial. Thus, the roots are 3,1,0. These are the eigenvalues. If I have equations,

(1-t)x + 2y = 0
1x + (2-t)y = 0
(1-t)z = 0,

and I plug in for t=0,1,3, I find for t=3 that eigenvectors are multiples of (1,1,0). For t=1, eigenvectors are multiples of (0,0,1). For t=0, eigenvectors are multiples of (-2,1,0). The matrix is diagonalizable because T has three linearly indep. eigenvectors.

Because these vectors are linearly independent, and because the number of vectors = dim(R3), these vectors span R3. Thus, R3 is the eigenspace of T (???)

How does that look?