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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.

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    $\begingroup$ The matrix $\begin{pmatrix} 5^2 & 2^2\\ 7^2 & 3^2 \end{pmatrix}$ has as eigenvalues $17\pm 2\sqrt{65}$ $\endgroup$ Sep 29, 2019 at 1:03
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    $\begingroup$ The radical should be $\sqrt{a^4 + 4 b^2 c^2 - 2 a^2 d^2 + d^4}$. $\endgroup$
    – YiFan
    Sep 29, 2019 at 1:05
  • $\begingroup$ Does this all look correct now? Thanks for pointing that out $\endgroup$
    – PMaynard
    Sep 29, 2019 at 1:08
  • $\begingroup$ @SebastianCor is $\sqrt{65}$ rational? $\endgroup$
    – Time4Tea
    Sep 29, 2019 at 1:10
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    $\begingroup$ Also, since the characteristic polynomial of a matrix is monic, any rational eigenvalues of such matrices (or any matrices with integer entries) in fact must be integral. $\endgroup$ Sep 30, 2019 at 0:48

3 Answers 3

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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.

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    $\begingroup$ Thanks! So I presume then there are probably 3x3 matrices that have irrational eigenvectors? Are they just really difficult to find? (is there a reason they are abundant in $3x3$ but hard to find in $2x2$? $\endgroup$
    – PMaynard
    Sep 29, 2019 at 1:15
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    $\begingroup$ @Time4Tea I simply asked Mathematica to find instances where the radical (in the question body) is equals to an integer, and $a,b,c,d$ are distinct. It returned the answer quite quickly. $\endgroup$
    – YiFan
    Sep 29, 2019 at 1:19
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    $\begingroup$ Ah, ok. Nice effort :) $\endgroup$
    – Time4Tea
    Sep 29, 2019 at 1:20
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    $\begingroup$ Haha I just found it while I was studying these matrices, I made the observation and then started trying to figure out why and couldn't find an explanation. Also I'm looking now to see if there is one with non-negative values? It seems there is probably some reason for why this happens too which I'd like to investigate. And if this occurs under a special condition. $\endgroup$
    – PMaynard
    Sep 29, 2019 at 1:22
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    $\begingroup$ Here is another with non-negative eigenvalues and no ones, $ \begin{pmatrix} 14^2 & 36^2 \\ 5^2 & 26^2 \end{pmatrix}$ $\endgroup$
    – PMaynard
    Sep 29, 2019 at 1:48
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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.

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    $\begingroup$ Thanks, and you're welcome! I wrote a short Maple script to find minimal $3 \times 3$ examples whose eigenvalues are rational (which turns out, I think, to be a more interesting question) and updated my post with some data. $\endgroup$ Sep 30, 2019 at 1:21
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    $\begingroup$ You're welcome. I'm reasonably sure that for any $m$ if we randomly assign distinct elements of $\{0^2, \ldots, N^2\}$ to the entries of an $m \times m$ matrix that the probability of at least one root of the characteristic polynomial being rational goes to $0$ as $N \to \infty$. If you're familiar with Galois theory, there's a related result about the probability of a Galois group of a polynomial of fixed degree with integer coefficients is $S_n$ goes to $1$ as we exhaust the set of integer polynomials in certain reasonable ways. $\endgroup$ Sep 30, 2019 at 2:39
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    $\begingroup$ Tweaking it just now, by the way, found the lexicographically first example (with entries $0^2, \ldots, 8^2$) $$\pmatrix{0^2&1^2&2^2\\4^2&6^2&8^2\\5^2&3^2&7^2},$$ which has rational eigenvalue $69$ and irrational eigenvalues $8 \pm 4 \sqrt{6}$. $\endgroup$ Sep 30, 2019 at 2:43
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    $\begingroup$ It was little trouble, I've added the code to my answer. I learned Galois Theory mostly from Dummit and Foote. I really like the exposition but it's a bit verbose for some tastes, I've heard. $\endgroup$ Sep 30, 2019 at 3:20
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    $\begingroup$ By the way, my code found that $5064$ of the $9!$ matrices of the above form have precisely one rational root. $\endgroup$ Sep 30, 2019 at 3:21
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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.

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