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Suppose $A \in M_{n\times n}(\Bbb R)$ has minimal polynomial $$(\lambda-1)^2(\lambda+1)^2$$ and $n\leq 6$. Find all possible Jordan normal forms.

If we denote the $n \times n$ Jordan block

$$J_\alpha (n) = \begin{bmatrix}\alpha&1&\cdots&\cdots\\0&\alpha&1&\cdots\\\vdots&\vdots\\0&0&\cdots&\alpha\end{bmatrix}$$

Then for the $n=6$ case: the characteristic polynomial is

$$(\lambda-1)^3(\lambda+1)^3$$

or

$$(\lambda-1)^4(\lambda+1)^2$$

or

$$(\lambda-1)^2(\lambda+1)^4$$

So are the possible Jordan normal forms the following?

$$\left\{ \begin{bmatrix}J_1(3)&O\\O&J_{-1}(3)\end{bmatrix}, \begin{bmatrix}J_{-1}(3)&O\\O&J_{1}(3)\end{bmatrix}, \begin{bmatrix}J_1(4)&O\\O&J_{-1}(2)\end{bmatrix}, \\ \begin{bmatrix}J_{-1}(2)&O\\O&J_{1}(4)\end{bmatrix}, \begin{bmatrix}J_1(2)&O\\O&J_{-1}(4)\end{bmatrix}, \begin{bmatrix}J_{-1}(4)&O\\O&J_{1}(2)\end{bmatrix} \right\}$$

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    $\begingroup$ I don't think you can have a $3\times 3$ (or bigger) Jordan block, e.g. $J_1(3)$, because this would make the minimal polynomial $(\lambda-1)^3(\lambda+1)^{\text{something}}$. You can have only $2\times 2$ and $1\times 1$ blocks, with at least one $2\times 2$ block for each of eigenvalues $-1$ and $1$. $\endgroup$ – user491874 Feb 6 '18 at 12:21
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    $\begingroup$ (Cont'd) So I think you can have either $J_1(2), J_1(2), J_{-1}(2)$, or $J_1(2), J_1(1), J_1(1), J_{-1}(2)$ or $J_1(2), J_1(1), J_{-1}(2), J_{-1}(1)$ or $J_1(2), J_{-1}(2), J_{-1}(1), J_{-1}(1)$ or $J_1(2), J_{-1}(2), J_{-1}(2)$. $\endgroup$ – user491874 Feb 6 '18 at 12:25
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Any of the Jordan blocks cannot be of size $3$ (or more) because it would then correspond to a factor $(\lambda\pm 1)^{3\text{ or more}}$ in the minimal polynomial. However, to get to the minimal polynomial with factors $(\lambda\pm 1)^2$, there must be at least one Jordan block of size 2. Therefore, the solutions for $n=6$ are:

$$\left[\begin{array}{cc|cc|cc}1&1&0&0&0&0\\0&1&0&0&0&0\\\hline0&0&1&1&0&0\\0&0&0&1&0&0\\\hline0&0&0&0&-1&1\\0&0&0&0&0&-1\end{array}\right]$$

$$\left[\begin{array}{cc|c|c|cc}1&1&0&0&0&0\\0&1&0&0&0&0\\\hline0&0&1&0&0&0\\\hline0&0&0&1&0&0\\\hline0&0&0&0&-1&1\\0&0&0&0&0&-1\end{array}\right]$$

$$\left[\begin{array}{cc|cc|cc}1&1&0&0&0&0\\0&1&0&0&0&0\\\hline0&0&-1&1&0&0\\0&0&0&-1&0&0\\\hline0&0&0&0&-1&1\\0&0&0&0&0&-1\end{array}\right]$$

$$\left[\begin{array}{cc|cc|c|c}1&1&0&0&0&0\\0&1&0&0&0&0\\\hline0&0&-1&1&0&0\\0&0&0&-1&0&0\\\hline0&0&0&0&-1&0\\\hline0&0&0&0&0&-1\end{array}\right]$$

$$\left[\begin{array}{cc|c|cc|c}1&1&0&0&0&0\\0&1&0&0&0&0\\\hline0&0&1&0&0&0\\\hline0&0&0&-1&1&0\\0&0&0&0&-1&0\\\hline0&0&0&0&0&-1\end{array}\right]$$

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  • $\begingroup$ Thanks! Very clear! $\endgroup$ – Eric Feb 6 '18 at 14:55
  • $\begingroup$ beautiful blocking, it makes it much easier to spot the blocks. $\endgroup$ – mathreadler Feb 7 '18 at 7:25

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