Finding matrices with polynomials of an exact form Might be a bad title here, but I am looking through some old prelim questions and am curious if someone may assist with this one:
Find a list of real matrices, as long as possible, such that:
1) The characteristic polynomial of each matrix $p(x) = (x-1)^{5}(x+1)$.
2) The minimal polynomial of each matrix is $m(x) = (x-1)^{2}(x+1)$.
3) No two matrices in the list are similar to each other.
Any tips? Is this leading to Jordan normal form?
 A: Do you understand the distinction between "characteristic polynomial" and "minimal polynomial"?  We can get the characteristic polynomial by subtracting, say, $\lambda$, from each number on the main diagonal and then taking the determinant which will be a polynomial in $\lambda$, the "characteristic polynomial".  It can be shown that every matrix makes its characteristic polynomial zero.
But, if the characteristic polynomial has repeated factors (so has repeated eigenvalues), we might be able to remove some of those factors getting a polynomial of lower degree that the matrix still makes 0.  That is the "minimal polynomial".  Yes, this has to do with the "Jordan Normal Form".  For example the matrices $A= \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}$ and $B= \begin{bmatrix}3 & 1 \\ 0 & 3\end{bmatrix}$ both has "characteristic polynomial" $(3- \lambda)^2$. 
Both matrices satisfy that polynomial: $3I- A=  \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}- \begin{bmatrix} 3 & 0 \\ 0 & 3\end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix}$ and $3I- B= \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}- \begin{bmatrix} 3 & 1 \\ 0 & 3\end{bmatrix}= \begin{bmatrix}0 & -1 \\ 0 & 0 \end{bmatrix}$ both of which square to give 0.  But $3I- A$ itself is the 0 matrix while $3I- B$ is not.  That is, the "minima polynomial" for A is $3- \lambda$ while the "minimal polynomial" for B is $(3- \lambda)^2$.
A: Hint:
See Frobenius normal form and similarity invariants. 
For any endomorphism $f$ of a vector space $E$ of dimension $n$ over a field $K$, $E$ can be seen as a finitely generated module over the ring of polynomials $K[X]$ through  $X\cdot u=f(u)$. It is a torsion module over this ring.
As $K[X]$ is a P. I. D., the $K[X]$-module $E$, by the Fundamental theorem of modules over P. I. D.s is a direct sum of cyclic submodules $\bigoplus_{i=1}^rE_i$, each $\;E_i$ isomorphic to $K[X]/(P_i)$, such that for all $i=1,\dots r-1, \enspace P_i\mid P_{i+1}$.
The sequence of polynomials $(P_1,\dots, P_r)$ characterises the endomorphism (or its matrix in a given base) up to similarity.
Furthermore, we know that $P_r$ is the minimal polynomial of the endomorphism $f$, and the product $P_1\dotsm P_r\;$  is its characteristic polynomial. Hence all you have to do is to find what are the possible steps to go from $P_r(X)=(X-1)^2(X+1)$ to $P_1$ with the above-mentioned constraints. Then, for each similarity invariants, you  can take as an example of a similarity class the block-matrix with each block equal to the companion matrix of the invariants.
A: These are $6\times 6$ matrices because the characteristic polynomial is a sixth degree polynomial. There are two eigenvalues: $\lambda=-1$ and $\lambda=1$. The Jordan blocks with eigenvalue $-1$ must all be order $1$. The Jordan blocks with eigenvalue $\lambda=1$ must have at least one $2$ block, but no larger.
Use $+$ to denote eigenvalues $1$ and use $-$ to denote eigenvalues $-1$. let $[\;\;]$ denote a grouping by block.
$$
\mbox{2 [++] blocks} \\
          [++][++][+][-] \\
          [++][++][-][-] \\
\mbox{1 [++] block} \\
          [++][+][+][+][-] \\
          [++][+][+][-][-] \\
          [++][+][-][-][-] \\
          [++][-][-][-][-]
$$
