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Given $T: V \to V$ a linear operator, and given $P_T(x) = x^2(x-1)^5(x+4)^4$ and $M_T(x) = x(x-1)^2(x+4)^3$. Find the possibilities of the Jordan form of $T$.

So I said this:

Eigenvalues $0,1,-4$. The sum of the sizes of the Jordan blocks for eigenvalue $0$ is 2, for eigenvalue $1$ is $5$, and for eigenvalue $-4$ is $4$. And then I said, the max block size of eigenvalue of $0$ is $1$, for eigenvalue $1$ is $2$, for eigenvalue $-4$ is $3$. So from this I figured out the following possibilities:

(I) For eigenvalue $0$: 2 blocks of size $1 \times 1$ each. For eigenvalue $-4$: one block of size $3 \times 3$ and one block of size $1 \times 1$. For eigenvalue $1$: two blocks $2 \times 2$ and one block $1 \times 1$.

(II) For eigenvalue $0$: 2 blocks of size $1 \times 1$ each. For eigenvalue $-4$: one block of size $3 \times 3$ and one block of size $1 \times 1$. For eigenvalue $1$: one block of size $2 \times 2$ and one block of size $3 \times 3$.

(III) For eigenvalue $0$: 2 blocks of size $1 \times 1$ each. For eigenvalue $-4$: one block of size $3 \times 3$ and one block of size $1 \times 1$. For eigenvalue $1$: one block $2 \times 2$, 3 blocks $1 \times 1$.

Is that correct or am I missing something?

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3 Answers

The exponent of $(x-\lambda)$ in the minimal polynomial gives the size of the largest Jordan block with diagonal entries $\lambda$. For $\lambda=1$ this exponent is $2$, so there are no $3\times3$ Jordan blocks for this eigenvalue, and option (II) is wrong. The other two options are the only possibilities.

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For all values of one-hundred-per-cent, there are $exactly$ 2 possible Jordan forms [your options (I) and (III)].

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up vote 0 down vote accepted

Apparently my answer was one-hundred-per-cent correct, I asked my professor about it.

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Only for rather small values of one-hundred-per-cent. –  Marc van Leeuwen Oct 10 '13 at 12:51
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