De-diagonalize a diagonal matrix back over a subfield from the extension field

Question

I think this is a pretty strange question because we usually want to diagonalize a matrix, not de-diagonalize it. However, this idea looks helpful in my work, so I like to ask. Let the field extension $K \subset L$ be a Galois extension, and $D$ be a diagonal matrix over $L$. My question is about (1) characterizing when $D$ can be written as a matrix over $K$ (up to a similarity transformation $T$ over $L$) and (2) how to do that explicitly, i.e. to find such a similarity transformation $T$. Necessarily, if it is possible, then the characteristic equation of $D$ has $K$-coefficients, but it splits over $L$, and the minimal polynomial of $D$ splits into distinct linear factors over $L$.

For example, consider the case of $\mathbb{C} = L$ and $K = \mathbb{R}$. We consider the matrix $$D = \begin{bmatrix} i & 0 & 0 & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & 0 & 0 & -i \end{bmatrix}. $$ Then, the characteristic polynomial is $(x-i)^2(x+i)^2 = (x^2 +1)^2$ and the minimal polynomial is $(x-i)(x+i) = x^2 + 1$. However, actually, we know that $$ P^{-1}A P =P^{-1}\begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{bmatrix} P = D, $$ where $$P= \begin{bmatrix} 0 & i & 0 & -i \\ 0 & 1 & 0 & 1 \\ i & 0 & -i & 0 \\ 1 & 0 & 1 & 0 \\ \end{bmatrix}. $$

However, this example can be obtained because we already know that the matrix $A$ can be diagonalized to $D$. I believe that this is an example or an exercise from many linear algebra textbooks.

I think that finding a rational form (over $K$) from the data of $D$ can be useful. In the previous example, the minimal polynomial over $\mathbb{R}$ is $x^2 +1$, so by using the invariant factors $x^2+1$, $x^2+1$, we can find the form of $A$. Each block in $A$ is exactly the companion matrix of $x^2 +1$. Moreover, I think the coincidence of the minimal polynomials over $\mathbb{R}$ and over $\mathbb{C}$ was also important in this example.

In fact, I need to apply this kind of idea to a function field and a radical extension of it, but I guess a general discussion might be useful for clarifying the situation.

Answer 1

This is possible iff the characteristic polynomial of $D$ has coefficients in $K$, or equivalently iff the diagonal entries of $D$ are invariant under the action of the Galois group $G = \text{Gal}(L/K)$ (when considered with multiplicity).

This is clearly a necessary condition. If it holds, consider the irreducible factorization $\prod f_i(t)^{m_i}$ of the characteristic polynomial over $K$. Then construct the matrix $A$ over $K$ which is block-diagonal with blocks given by $m_i$ copies of the companion matrix of $f_i(t)$. $A$ is, by construction, diagonalizable over $L$ with the same eigenvalues as $D$, and hence is conjugate to $D$ over $L$. (Edit: We don't actually need $L/K$ to be a Galois extension here, it suffices to require that it be separable to conclude that $A$ is diagonalizable.)

To find an explicit matrix that conjugates $A$ to $D$ it suffices to work one block at a time, so we can reduce WLOG to the case that the characteristic polynomial is a single irreducible polynomial $f(t) \in K[t]$ which splits over $L$ (equivalently, to the case that the diagonal entries of $D$ are a single Galois orbit). Writing $K_f = K[\alpha]/f(\alpha)$, the companion matrix of $f$ can be understood abstractly as the matrix of $\alpha$ acting by left multiplication on $K_f$ in the basis $\{ 1, \alpha, \dots \alpha^{\deg f - 1} \}$. We would like to find a different basis, namely a basis of eigenvectors of multiplication by $\alpha$, after extending scalars to $L$. This means we want to understand the tensor product

$$L_f \cong K_f \otimes_K L \cong K[\alpha]/f(\alpha) \otimes_K L \cong L[\alpha]/f(\alpha)$$

as an $L[\alpha]$-module. By hypothesis, $f(t) = \prod_i (t - \lambda_i)$ splits over $L$, so the Chinese remainder theorem produces an isomorphism

$$L_f \cong \prod_i L[\alpha]/(\alpha - \lambda_i)$$

where each factor is just isomorphic to $L$ but we write it as $L[\alpha]/(\alpha - \lambda_i)$ to emphasize what the map from the LHS to the RHS is: it is given by evaluating an element $g(\alpha) \in L[\alpha]$ on each of the roots $\lambda_i$ of $f$.

The eigenvectors of $\alpha$ acting by left multiplication on $L_f$ are given (up to scale) by the preimages $e_i(\alpha)$ of the idempotents in $\prod_i L[\alpha]/(\alpha - \lambda_i)$ whose $i^{th}$ component is equal to $1$ and whose other components are equal to zero. We can write these down explicitly in terms of Lagrange interpolating polynomials: namely, they are given by

$$e_i(\alpha) = \frac{\prod_{j \neq i} (\alpha - \lambda_j)}{\prod_{j \neq i} (\lambda_i - \lambda_j)}.$$

By construction these satisfy $e_i(\lambda_j) = \delta_{ij}$, and they also satisfy $(\alpha - \lambda_i) e_i(\alpha) \equiv 0 \bmod f(\alpha)$, hence are eigenvectors of $\alpha$ acting on $L_f$ with eigenvalue $\lambda_i$, as desired. Expanding the above out in terms of $\alpha$ produces the desired change-of-basis matrix $P$ (or its inverse, I'm not sure), although personally I think working with these expressions directly is nicer than thinking in terms of matrices. An alternative expression for $e_i$ directly in terms of $f$ is

$$e_i(\alpha) = \frac{f(\alpha)}{(\alpha - \lambda_i) f'(\lambda_i)}$$

where this division needs to be understood as happening in $L[\alpha]$ before passing to the further quotient by $f(\alpha)$. This also came up recently here.