When are the eigenvalues of a matrix containing all squared elements irrational/rational?

Question

Forgive me in advance if any of this is trivial. After looking at many 2x2 matrices it seems that if all of the elements in matrix are unique squared integers then the eigenvalues are irrational. So I tried to investigate this:

$\det \begin{pmatrix} \lambda -a^2 & b^2\\ c^2& \lambda -d^2\end{pmatrix}= \lambda^2 -(a^2+d^2)\lambda + (a^2d^2-c^2b^2)$

after applying the quadratic formula this gives a radical of,

$\sqrt{a^4+4b^2c^2-2a^2d^2+d^4}$

If the stated observation is true, is there a way to show that this is irrational? Furthermore it looks like on the surface that for 3x3 matrices the eigenvalues for a matrix containing all unique squared entries that the eigenvalues will also be irrational. Are either of these statements true? Is there a generalization of this for an nxn matrix?

Edit: I'm not entirely sure I derived the radical correctly, but I'd still like to have some direction on the questions above also I'd like to examine cases where the eigenvalue is not zero

Example: \begin{pmatrix} 2^2 & 4^2\\ 3^2 & 6^2 \end{pmatrix} has eigenvalues 40 and 0.

Edit 2: still looking for rational eigenvalues of a $3x3$ have been with imposed restrictions and nonzero eigenvalues/entries.

Answer 1

The claim is not true. The matrix $$\begin{bmatrix}1^2&36^2\\5^2&26^2\end{bmatrix}$$ has eigenvalues $721$ and $-44$, which are evidently rational.

Answer 2

If $b = 0$ or $c = 0$, then the matrix is triangular and so the eigenvalues are just the diagonal entries $a^2, d^2$ (which in particular are rational). So, any of the $12$ choices with $\{a, b, c, d\} = \{0, 1, 2, 3\}$ and either $b = 0$ or $c = 0$ gives a solution, and these examples evidently minimize $\max\{a, b, c, d\}$ among nonnegative solutions.

If we exclude $0 \in \{a, b, c, d\}$ to avoid these trivial solutions, the minimal solutions and their eigenvalues are \begin{array}{cc}\hline (a, b, c, d) & \lambda \\ \hline(1, 2, 5, 4) & -4, 21 \\ (4, 2, 3, 5) & 13, 28 \\ \hline \end{array} and the six examples obtained from these using the evident symmetries $a \leftrightarrow d$ and $b \leftrightarrow c$.

For the $3 \times 3$ case already the matrix $$\pmatrix{0^2 & 1^2 & 2^2 \\ 3^2 & 4^2 & 5^2 \\ 6^2 & 7^2 & 8^2}$$ has irrational eigenvalues: Its characteristic polynomial, $c(t) = t^3 - 80 t^2 - 354 t + 216$, has positive discriminant, so it has three real roots. On the other hand, $c(t) \equiv t^3 + t + 1 \pmod 5$, but this latter polynomial has no roots modulo $5$, hence $c(t)$ is irreducible over $\Bbb Q$, that is, its three real roots are irrational.

In a sense that can be made precise, most rational polynomials do not have all roots rational, and I see little reason to expect that the characteristic polynomials of matrices with (distinct) square entries would be special in this regard, so it is perhaps more interesting to ask for an example whose eigenvalues are rational (and hence integral). A quick Maple script (transcribed below) finds many examples with entries $0^2, \ldots, 8^2$. The first of these lexicographically is $$\pmatrix{0^2 & 2^2 & 3^2 \\ 5^2 & 8^2 & 1^2 \\ 7^2 & 6^2 & 4^2} , \quad \textrm{which has eigenvalues} \quad {-13}, 24, 69 .$$ It is rare even among the matrices with entries $0^2, \ldots, 8^2$ to have all rational eigenvalues: It only happens for $252$ of the $9! = 362880$ cases, $180$ of which have no zero eigenvalues.

restart;
with(combinat): with(LinearAlgebra):
m := 3;
N := 9;

for numberSet in choose(N, m^2) do
    shifted := map(U -> U - 1, numberSet);
    print([shifted]);
    for ordering in permute(shifted) do
        map(U -> U^2, ordering);
        A := convert([seq(%[((i - 1) * m + 1)..(i * m)], i=1..m)], Matrix);
        c := CharacteristicPolynomial(A, t);
        if (convert(map(degree, map(U -> U[1], factors(c)[2]), t), set) = {1}) then
            print(ordering, A, solve(c));
        fi:
    od:
od:

Here the constant $m$ is the matrix size, $[0, \ldots, N - 1]$ is the range from which the squared numbers are chosen.

Answer 3

In conclusion for anyone who comes across this later, there are many cases where eigenvalues are rational for $2x2$ matrices containing only distinct squared integers.The case were the eigenvalues are rational just appear less frequently. Looking at the radical from the question this becomes apparent:

$\sqrt{a^4+4b^2c^2-2a^2d^2+d^4}$

The reason they are not as easy to find is because in order for them to be rational, $a^4+4b^2c^2-2a^2d^2+d^4$ must be a squared number and this happens less frequently. The $3x3$ case has more or less the same explanation/result.