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A classic problem in linear algebra is to determine if two matrices $A,B\in M_n(\mathbb{F})$ are similar one to another. When $\mathbb{F}=\bar{\mathbb{F}}$, we know that $A,B$ are similar if and only if they have the same Jordan form.

What about the the case where $\mathbb{F}\neq \bar{\mathbb{F}}$? In this case I saw that there is a theorem saying that for two matrices $A,B\in M_n(\mathbb{F})$, if they are similar over $\bar{\mathbb{F}}$ then they are similar over $\mathbb{F}$.

However, I can't find any "natural" proof for that. I want to ask if someone know such a proof?

azimut
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Ofir Schnabel
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2 Answers2

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The argument of orangeskid can be generalized to all cases where the base field is infinite. Unfortunately, for the general case my impression is that no such basic proof is known.

Typically, the statement is given as a corollary of the rational canonical form, which is a normal form for square matrices under the equivalence notion of similarity. In contrast to the Jordan normal form, the rational canonical form works over any field, no matter if algebraically closed or not. The rational canonical form results as an application of the classification of finitely generated modules over principal ideal rings, which should be found in most book on abstract algebra.

The argument is the following:

If $L$ is a field and $A,B$ are square matrices with entries in $L$, then $A$ and $B$ are similar if and only if they have the same rational canonical form. The rational canonical form is determined by the invariant factors of the matrices, which are polynomials over $L$.

If the entries of $A$ are contained in a subfield of $K$, then the coefficients of the invariant factors are contained in $K$, too. This shows that in this case, $A$ and $B$ are similar over $L$ if and only if they are similar over $K$.

Another way to see it: The rational canonical form has the nice property of being invariant under the transition to extension fields. Actually, this is the meaning of the word "rational" in this context.

azimut
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  • Thanks, I will check what are these invariant factors. Where it is proved that two matrices are similar if and only if they have the same invariant factors? – Ofir Schnabel Jun 25 '15 at 09:35
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    @OfirSchnabel: I've added a bit more info on the theoretical background. Just open your favorite book on abstract algebra and look for the classification of finitely generated modules over principal ideal rings. The rational canonical form should follow closely behind. – azimut Jun 25 '15 at 11:47
  • Thanks a lot. I will look for it. It seems however, that this proof is not for linear algebra students as I wanted, but I will read it for myself. – Ofir Schnabel Jun 25 '15 at 13:21
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    @OfirSchnabel: It depends. There are different possibilities for the proof of the Jordan normal form in a linear algebra course. Sometimes, the full classification of modules over PIDs is done, having the Jordan normal form as a corollary. (The time needed to do this is comparable with the "classical" proof. However, it is even more abstract and harder to follow, which makes it a debatable option for a freshman course.) In this case, all the needed groundwork is there. – azimut Jun 25 '15 at 13:28
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In a linear algebra class you can show that if $A$, $B \in M_n(\mathbb{R})$ are similar in $M_n(\mathbb{C})$, then they are similar. For there exists $X + i Y \in M_n(\mathbb{C})$ with $$A (X+ iY) = (X+iY) B$$ and $\det (X+iY) \ne 0$. We get $AX= XB$, $AY= YB$ so $A(X+tY)=(X+tY)B$ for any $t\in \mathbb{R}$. You only have to choose $t$ so that $\det(X+tY)\ne 0$. This is possible, because the polynomial $t\mapsto \det(X+tY)$ is not $0$, for taking a nonzero value at the complex value $t=i$.

azimut
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orangeskid
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    @No worries. Indeed $K$ infinite is needed. As for the extension $\tilde K /K$, since we only use finitely many elements ( the entries of the conjugating matrix) we always have them in a finite dimensional space. – orangeskid Feb 16 '15 at 08:26