I heard the following theorem from our textbook:

Given $A,B$ are two commuting ($AB=BA$) real normal matrices. There's some real orthogonal matrix $P$ such that $P^{-1}AP$, $P^{-1}BP$ are canonical forms.

The canonical form of a real normal matrix is a block diagonal whose diagonal blocks are either $\lambda\in\mathbb R$ or \begin{bmatrix}a&-b\\b&a\end{bmatrix}

It could be proved by induction on the size of the matrices. It's well-known that there's a common (complex) eigenvector $v$. If $v$ is a real vector (or a scalar of a real one), then the reduction is obvious. If $v=x+iy$, where $x,y$ are linear independent, since $A$ is orthogonally similar to a canonical form, it's not so hard to show that $x^tx=y^ty$ and $x^ty=y^tx=0$, and $\operatorname{span}\{x,y\}$ is $A$(and $B$)-invariant.

However, I'm not satisfied with the proof above. It seems that the paradigm could be generalized as follow:

Suppose $A,B$ are semi-simple matrices over a number field $K$ which could be simultaneously diagonalized over $\mathbb C$, then there's some matrix $P$ over $K$ such that $P^{-1}AP$, $P^{-1}BP$ are block diagonal matrices, the characteristic polynomial of whose diagonal blocks are irreducible.

Even more, I wonder whether we could stipulate these blocks to be Frobenius normal forms. I don't know, however.

Any proof (or disproof) of the preceding statement or any generalization? Thanks!


  1. Semisimple matrices are matrices for semisimple operators. For example, if the number field $K=\mathbb C$ or another algebraically closed field, semisimple matrices are just diagonalizable matrices, therefore the statement is true when $K$ is algebraically closed.

1 Answer 1


Your generalization is false. For instance, let $K=\mathbb{Q}$ and $V=\mathbb{Q}(\sqrt{2},\sqrt{3})$, with $A$ acting on $V$ as multiplication by $\sqrt{2}$ and $B$ acting on $V$ as multiplication by $\sqrt{3}$. If a simultaneous block-diagonalization of the sort you ask for exists, then all the blocks must be $2\times 2$, since the minimal polynomials of $A$ and $B$ are both irreducible of degree $2$. It follows that the span of the first two basis vectors must be invariant under both $A$ and $B$. But $V$ is irreducible with respect to $A$ and $B$--the only nontrivial subspace of $V$ which is invariant under both $A$ and $B$ is $V$ itself (since such a subspace is then invariant under multiplication by any element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$). So no such simultaneous block-diagonalization can exist.

In general, what will happen is that you can decompose your vector space $V$ as a direct sum of subspaces $V_i$ which have the form of the example above: each $V_i$ is a finite separable field extension of $K$, with $A$ and $B$ corresponding to elements of the field which generate the field over $K$ and act by multiplication. (Proof sketch: If $f$ and $g$ are the minimal polynomials of $A$ and $B$, then $V$ is a $K[x,y]/(f(x),g(y))$-module via $A$ and $B$. Since $A$ and $B$ are semisimple, $f$ and $g$ are separable and it follows that the ring $K[x,y]/(f(x),g(y))$ is semisimple, so any module over it is a direct sum of simple modules which correspond to field quotients of $K[x,y]/(f(x),g(y))$.)

In the case $K=\mathbb{R}$, these field extensions can only be $\mathbb{R}$ or $\mathbb{C}$, and in particular each of $A$ or $B$ is either just an element of $\mathbb{R}$ or else generates the entire field. It is then easy to see you can choose a basis for $V_i$ such that both $A$ and $B$ have the form you ask for. But in general, you can do no such thing, as the counterexample above shows.


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