Let $A \in \mathbb{M_6(R)}$ be a real matrix with the minimal polynomial $m_A(x) = (x-1)^2(x-2)$ and characteristic polynomial $c_A(x) = (x-1)^4(x-2)^2$. Find all the possible Jordan canonical forms of A.

Can someone explain to me how does the forms of minimal polynomials and characteristic polynomials tell me about its Jordan forms. Thanks.


1 Answer 1


Let $T$ be the linear operator on $\mathbb{R}^6$given by $A$. Then by structure theorem $V \simeq \mathbb{R}[x]/(f_1(x)) \oplus \cdots \oplus \mathbb{R}[x]/(f_n(x))$ as a $\mathbb{R}[x]$ module where $(f_1(x)) \supset (f_2(x)) \cdots \supset(f_n(x))$ (containment of ideals). Also $char(T)=f_1(x)\cdots f_n(x)$ and $min(T)=f_n(x)$.

Now, given minimum polynomial and characteristic polynomial we just want the number of possibilities of $n $ and $f_i$'s and each such choice will give us a Jordan decomposition(Each $f_i$ factors completely as it is a factor of $f_n$ and $f_n$ factors completely). Conversely as well given a Jordan decomposition of a operator with minimal and characteristic polynomial as given in the problem, it determines the ideals $(f_i(x))$.

The only possibilities are $n=2$ and $f_1(x)=(x-1)^2(x-2)$, $f_2(x)=(x-1)^2(x-2)$ and $n=3$ and $f_1(x)= (x-1)$, $f_2(x)= (x-1)(x-2)$, $f_3(x)= (x-1)^2(x-2)$. The first one has $2$ blocks of size $2$ for eigenvalue $1$ and $2$ blocks of size $1$ for eigenvalue $2$. The second one has $2$ blocks of size $1$ and $1$ block of size $2$ for eigenvalue $1$ and $2$ blocks of size $1$ for eigenvalue $2$.

Hope it helps.


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