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**Deveno** what is dim(ker(A-I))? if we row-reduce A-I, we find it has rank 2. therefore dim(R^3) = rank(A-I) + dim(ker(A-I)):

3 = 2 + dim(ker(A-I)), so dim(ker(A-I)) = 1.

OR...let's find ker(A-I) explicitly:

$\displaystyle \begin{bmatrix}0&-3&3\\-2&-7&13\\-1&-4&7 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}$

right away, we see that x2 = x3.

from either 2 of the remaining equations, we get x1 = 3x2.

so {(3,1,1)} is a basis for A-I. so the Jordan form for A has just one Jordan block, in other words the entire superdiagonal is 1's.

now...you agree that if J is the jordan form for A, that there is some matrix P with:

$\displaystyle A = PJP^{-1}$ right? so

$\displaystyle A - I = PJP^{-1} - I = PJP^{-1} - PIP^{-1} = P(J-I)P^{-1}$

what is the nilpotency of J-I? just look at it...it's obviously 2! J-I =

$\displaystyle \begin{bmatrix}0&1&0\\0&0&1\\0&0&0 \end{bmatrix}$

(J-I)^2 will raise the superdiagonal to the upper-right-hand corner, so (J-I)^2 won't be 0, so (A-I)^2 won't be 0.