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I have to prove the following result:

Let $A, B$ be two $n \times n$ matrices over the field $\sf F$ and $A,B$ have the same characteristic and minimal polynomials. If no eigenvalue has algebraic multiplicity greater than $3,$ then $A$ and $B$ are similar.

I have to use the following result:

If $A, B$ are two $3 \times 3$ nilpotent matrices, then $A, B$ are similar if and only if they have same minimal polynomial.

Please suggest how to proceed.

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Here's a sketch...

If $A$ and $B$ are similar, then $A=PBP^{-1}$ and so $A^k=(PBP^{-1})^k=PB^kP^{-1}$ and hence $f(A)=0$ if and only if $f(B)=0$. Thus similar matrices share the same minimal polynomial.

For the other direction. Suppose $A$ and $B$ share the same minimal polynomial and let $\lambda$ be one their (common) eigenvalues. Since the algebraic multiplicity of $\lambda$ is no more than $3$, there are only a few possibilities for the corresponding Jordan block(s) (in the Jordan form of $A$ or $B$). If the multiplicity is exact $3$ the possibilities are:

$$ J \qquad = \qquad \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}, \qquad \begin{bmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}, \qquad \mbox{or} \qquad \begin{bmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{bmatrix}$$

Notice that the first block is annihilated by $x-\lambda$, the second by $(x-\lambda)^2$, and the third by $(x-\lambda)^3$. Thus the minimal polynomial informs you as to which type of blocks appear. Thus since the matrices share the same Jordan form, they must be similar.

Notice the necessity of the hypothesis "algebraic multiplicity of $3$ or less". If you allow multiplicity $4$ you could have:

$$ \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{bmatrix} \qquad \mbox{and} \qquad \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{bmatrix} $$

Both of the above are annihilated by $(x-\lambda)^2$ but they are not similar (one has 2 blocks and so has 2 indep. eigenvectors and the other has 3 blocks and so has 3 indep. eigenvectors).

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  • $\begingroup$ Many thanks for the detailed reply. $\endgroup$ – preeti Mar 12 '12 at 14:55
  • $\begingroup$ My text book says that we have to use the mentioned result to prove the theorem. I am sorry to bother you again. It seems odd as result only holds for 3 X 3 nilpotent matrices. Please suggest. $\endgroup$ – preeti Apr 19 '12 at 21:25

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