# Jordan form and its elements

I am confused by the following question...

Find a matrix $A \in R^{3\times3}$, such that the minimal polynomial of $A$ is $(\lambda - 1)^2(\lambda + 2)$ and $\forall p,q \in \left\{ 1, 2, 3 \right\}$, $[A]_{pq} \in Z - \left\{0\right\}$ and $|[A]_{pq}| \leq 5$

The first thought that came to me is to construct the dot diagram by the minimal polynomial then write down the Jordan canoncial form, like this $$\begin{bmatrix} 1 & * & * \\ 1 & 1 & * \\ * & * & -2 \\ \end{bmatrix}$$

In my opinion, it should fill $0$ to those $*$; however, I don't know how to fill in with other integers.

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I think your second term should be $(\lambda+2)$ and not $(\lambda - 1)$. – Calvin Lin Jan 31 '13 at 0:58
I changed $3x3$ to $3\times 3$. That is standard TeX usage. – Michael Hardy Jan 31 '13 at 1:16
I caught Calvin's suggested edit. – RussH Jan 31 '13 at 1:20

Filling in the $\ast$ with $0$'s will give you the Jordan normal form of the matrix you're looking for. Then you just need to conjugate it (which does not change the minimal polynomial) into a matrix with the non-zero entries that you want. For example, conjugating by

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

does the trick.

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Thanks! But I have no idea what conjugate means...Find its conjugate transpose or what? – lucasKoFromTW Jan 31 '13 at 9:48
If $A$ is invertable then we way that $ABA^{-1}$ is the matrix $B$ conjugated by $A$. – Jim Jan 31 '13 at 16:24

A way to get an example regardless of what entry you wish in the $3,3$ position (and its negative in the linear factor of the characteristic polynomial) is to write down the Jordan Canonical Form including $0$ where you have your stars. That is set $$\mathbf{J}= \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & -2 \\ \end{bmatrix}$$ (or $1$ in the $3,3$ slot if you replace the last factor $(\lambda -1)$ in the characteristic polynomial by $(\lambda -2)$.

Now all you have to do is produce a nice $3\times 3$ matrix with integer entries that has an inverse with integer entries. I picked $$\mathbf {P} = \begin{bmatrix} 2 & 3 & 0 \\ 1 & 1 & 0 \\ 2 & 3 & 1 \\ \end{bmatrix}$$ and computed $\mathbf{P}^{-1}\mathbf{JP}$. No good--zeros in the last column. Pick another matrix $\mathbf{P_1}$ and compute $(\mathbf{PP_1})^{-1}\mathbf{JPP_1}$. Bingo! \begin{bmatrix} 7 & 9 & -18 \\ -24 & -29 & 54 \\ -10 & -12 & 22 \\ \end{bmatrix} works (unless I copied it wrong from my CAS). And it took much less time then writing this. Now for you--why did it work and how did I know the inverse of $\mathbf{P}$ would have all integer entries?

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Thanks! You're right! I do curious how you pick $P$ and the inverse of it. – lucasKoFromTW Jan 31 '13 at 9:46
Any integer matrix can be diagonalized by elementary row and column operations with entries in $\mathbf Z$ using the Euclidean algorithm. That is how the basis theorem for f.g. modules over a Euclidean domain is derived. The product of those entries is the determinant since all the elementary row operations have determinant 1. So P would have inverse with all integral entries iff it has determinant 1, and the best way to do that for me was to make P block lower triangular to eliminate 0 in the bottom row. Repeating with an upper block triangular matrix to clear the 3rd column. Use CAS. – Barbara Osofsky Jan 31 '13 at 14:56
By the way, I missed the requirement about values less than 5. – Barbara Osofsky Feb 1 '13 at 6:54