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The following questions is a review problem for qualifying material on Jordan normal forms and I am having a little trouble understanding the terminonology and using the fact we are given coprime polynomials corresponding to cyclic blocks.

Let $M$ be a block diagonal matrix over a field, consisting of two cyclic blocks whose characteristic polynomials are have a gcd of one.

How do we prove that it is possible to select a basis so the matrix becomes one cyclic block?

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3 Answers 3

up vote 2 down vote accepted

If you are familiar with the Frobenius normal form it is quite easy to prove the statement.

First we compute the Smith form of the characteristic matrix of $A$ (where $A$ is your matrix). Since we have two cyclic blocks this is basically the Smith form of a $2 \times 2$ diagonal matrix which has entries $f_1, f_2$ where those are the characteristic (and minimal) polynomials of your blocks. $f_1$ and $f_2$ are coprime hence your Smith form is $diag(1,...,1,f_1 \cdot f_2)$. Therefore the Frobenius form of your matrix consists of one cyclic block.

This is probably not exactly the desired answer since it does not use the Jordan form but it is a proof in the context of the various normal forms and it is often helpful to consider different forms to prove different statements.

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My guess would be that a cyclic block is one that is (or maybe is similar to) a companion matrix. This means the corresponding subspace is a cyclic $K[X]$-module, with $X$ acting as the given endomorphism $\phi$ (and also that this $K[X]$-module is a direct factor): for some vector its $\phi$-images span the subspace. To answer you question, one could take the sum of two such generating vectors for the blocks, just like two cyclic subgroups of coprime order in an abelian group have a (direct) sum that is again cyclic.

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Let $M$ be an $n$ by $n$ matrix with coefficient in a field $K$; view $K^n$ as a $K[X]$-module, where the indeterminate $X$ acts as $M$; and let $f,g\in K[X]$ be coprime polynomials.

Then the Chinese Remainder Theorem tells us that there is a $K[X]$-algebra isomorphism $$ \frac{K[X]}{(fg)}\ \overset{\sim}{\to}\ \frac{K[X]}{(f)}\times\frac{K[X]}{(g)}\quad. $$

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