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?