# Finding the minimal polynomial of a $A$, given a diagonal matrix equivalent to $XI_{10}-A$

Let $A\in M_{10}(\mathbb{C})$. We know that $XI_{10}-A$ is equivalent(By finite number of operations on the rows) to the diagonal matrix: $$C=\begin{pmatrix} 1& & & & & & & & & \\ & 1 & & & & & & & & \\ & &x & & & & & & & \\ & & &x+2 & & & & & & \\ & & & &x^2 & & & & & \\ & & & & &x+2 & & & & \\ & & & & & & (x^2+4)x & & & \\ & & & & & & & (x+2)^2 & & \\ & & & & & & & &1 & \\ & & & & & & & & & 1 \end{pmatrix}.$$ (the blank entries are $0$).

I need to find the minimal polynomial and Jordan form of $A$.

So I know know that the characteristic polynomial of A is: $F_A(x)= x^4(x+2)^4(x^2+4)$, and I'm having a hard realize what I can know in addition, to solve the problem. any hints? thanks!

• It seems to me there's something wrong here. The dimension of the generalized eigenspace $(X I_{10} - A)^4$ should be $4$ (the algebraic multiplicity of $0$), whereas here is $3$. – Giacomo d'Antonio Aug 15 '11 at 8:53
• @Arturo: It's a pity you deleted our chat. I really meant that. – user6163 Aug 16 '11 at 19:56
• If you want, ask a moderator to copy it onto the meta site. Have a good life. – Arturo Magidin Aug 16 '11 at 19:58
• You too. think of what I said. – user6163 Aug 16 '11 at 19:59

## 1 Answer

Let $K$ be a field, $X$ an indeterminate, and $V$ a finite dimensional $K[X]$-module.

Recall that the multiset of elementary divisors of $V$ is the multiset $$f_1^{n(1,1)},\dots,f_1^{n(1,k(1))},$$ $$\vdots$$ $$f_r^{n(r,1)},\dots,f_r^{n(r,k(r))},$$ defined by the following conditions: the $f_i$ are distinct monic irreducible polynomials, the $n(i,j)$ satisfy $$n(i,1)\ge\cdots\ge n(i,k(i)),$$ and we have $$V\simeq\bigoplus_{i,j}\frac{K[X]}{\big(f_i^{n(i,j)}\big)}\quad.$$

The characteristic polynomial is the product of the elementary divisors, whereas the minimal polynomial is the product of the $f_i^{n(i,1)}$.

If $W$ is also a finite dimensional $K[X]$-module, then the multiset of elementary divisors of $V\oplus W$ is the "disjoint union" (in the obvious sense) of the multisets of elementary divisors of $V$ and $W$.

Some diagonal entries of Nir's matrix are equal to 1, and the others are $$X,\quad X+2,\quad X^2,\quad X+2,\quad (X+2i)(X-2i)X,\quad (X+2)^2.$$ The multiset of elementary divisors is thus $$\begin{matrix} (X+2)^2,&X+2,&X+2,\\ \\ X^2,&X,&X,\\ \\ X+2i,&&\\ \\ X-2i.&& \end{matrix}$$ The characteristic polynomial is $$(X+2)^4X^4(X+2i)(X-2i).$$ The minimal polynomial is $$(X+2)^2X^2(X+2i)(X-2i).$$ The Jordan blocks $J(\lambda,n)$ are $$\begin{matrix} J(-2,2),&J(-2,1),&J(-2,1),\\ \\ J(0,2),&J(0,1),&J(0,1),\\ \\ J(-2i,1),&&\\ \\ J(2i,1).&& \end{matrix}$$

Let's compute the invariant factors (even if it wasn't required). In the array of elementary divisors, there are four blank entries. We put a $1$ in each of them: $$\begin{matrix} (X+2)^2,&X+2,&X+2,\\ \\ X^2,&X,&X,\\ \\ X+2i,&1,&1,\\ \\ X-2i,&1,&1. \end{matrix}$$ We get the first invariant factor by multiplying together the polynomials in the first row, and so on. Thus the invariant factors are $$(X+2)^2X^2(X+2i)(X-2i),\quad (X+2)X,\quad (X+2)X.$$

[Reference: Bourbaki, Algèbre, VII.4.8.]