If a matrix over the field $\mathbb R$ has as elementary divisors: $x-4$, $x^2 + 2$, does it then admit a Jordan canonical form?

Am I right in thinking that a matrix has a Jordan canonical form only when its elementary divisors are all of the form $(x-a)^n$?

sorry if my question is too basic. I am learning these stuff on my own and I am having some difficulty understanding them


1 Answer 1


A matrix over any field $F$ has a JCF if it splits over $F$, i.e. the elementary divisors of the matrix are of the form $(x-a)^n$. Alternatively, if $P[x]$ is a polynomial of degree $n$ over a field $F$, then $P[x] = 0$ has exactly $n$ roots over $F$.

So a matrix over $\mathbb{R}$ does not always have a JCF since $\mathbb{R}$ is not algebraically closed ($x^2 + 2$ does not split over $\mathbb{R}$ for example). However, a matrix over $\mathbb{C}$ (which is algebraically closed) always has a JCF (following from the example, $x^2 + 2$ does split into $(x - i\sqrt{2})(x + i\sqrt{2})$ over $\mathbb{C}$).

The rest of my answer comes from my own answer that I've posted somewhere else.

Furthermore, if $A$ is a square matrix with entries in a field $F$ and $F$ is an algebraically closed field (so the characteristic equation $c_A(x)$ splits over $F$), then $A$ has a Jordan normal form.

If $F$ is not algebraically closed, then there exists a field extension $K$ > $F$ such that $c_A(x)$ does split over $K$.


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