Product of eigenvalues over a finite field A have a matrix whose characteristic polynomial has 1 value over the ground field and 4 values over extension field. Let me call this matrix as P5. If I construct P25 as P5⨂P5, i.e, P25=P5⨂P5.
I notice that P25 characteristic polynomial has 9 eigenvalues over GF(5) and 16 in an extension field. But

Note that the product of eigenvalues by eigenvalues give us half of the values over ground field and other half in extension field.
This way I can predict that P25 has 1+8 = 9 values in the ground Field and 4+4+8=16 values over extension field.
The doubt is how can I prove that ExE produces half over Basis Field and half over Extension Field?? I tested with various matrices and 100% prediction was exact.
 A: Making an educated guess about what the asker is seeing. Consider the following matrix
$$
P=\left(
\begin{array}{ccccc}
 1 & 0 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 1 \\
 0 & -1 & 0 & -1 & -4 \\
\end{array}
\right)
$$
with entries viewed as elements of the field $GF(5)$. Because this matrix has a diagonal $1\times1$ block in the upper left, and the other diagonal $4\times4$ block is the companion matrix of the quartic
$$
f(x)=x^4+4x^3+x^2+1,
$$
we see that the characteristic polynomial of $P$ is
$\chi_P(x)=(x-1)f(x).$
I selected the polynomial $f(x)$ to be an irreducible factor of $(x^{39}-1)/(x^{13}-1)$ (in other words it is of exponent $39$). Therefore the zeros of $f(x)$ are roots of unity $\beta, \beta^5, \beta^{25},\beta^{125}=\beta^{8+3\cdot39}=\beta^8$ of order $39$.
The eigenvalues of $P\otimes P$ are then the pairwise products of eigenvalues of $P$. In other words, $1$, $\beta^{5^i}, i=0,1,2,3,$ (each with multiplicity two) and $\lambda_{i,j}:=\beta^{5^i+5^j}, 0\le i,j\le3$.
I claim that of those eigenvalues only $1$ belongs to the base field $GF(5)$.
Because $39$ is a factor of $624=5^4-1$ but is not a factor of $5^2-1$, the element $\beta$ generates the field $GF(5^4)$. The power $\beta^{13}$ is a cubic root of unity and thus belongs to $GF(5^2)$. 
We can conclude that


*

*The second group of eigenvalues, $\beta^{5^i}$ are all in $GF(5^4)\setminus GF(5^2)$.

*The eigenvalue $\lambda_{0,0}=\beta^2$, as well as its conjugates $\lambda_{i,i},i=1,2,3$, is of order $39$, and thus also in $GF(5^4)\setminus GF(5^2)$.

*The double eigenvalue $\lambda_{1,0}=\lambda_{0,1}=\beta^6$ is of order $13$, and therefore belongs to $GF(5^4)\setminus GF(5^2)$. The same applies to the conjugates $\lambda_{i,j}$ with $i\equiv j\pm1\pmod4$.

*The double eigenvalue $\lambda_{2,0}=\lambda_{0,2}=\beta^{26}$ is of order $3$, and therefor belongs to $GF(5^2)\setminus GF(5)$. The same applies to the conjugates $\lambda_{i,j}$ with $i\equiv j+2\pmod4$.


So this seems to disprove your conjecture. Only a single eigenvalue in the base field. But:


*

*Your conjecture will hold, if the characteristic polynomial of $P$ has two irreducible quadratic factors. The product of the zeros of such a factor is in $GF(5)$.

*Even with an irreducible quartic factor you may get some extra eigenvalues in $GF(5)$. For example if a primitive root of unity of order thirteen $\gamma=\beta^3$ is an eigenvalue of $P$, then this gives as eigenvalues of $P\otimes P$ elements like $\gamma\cdot\gamma^{5^2}=1\in GF(5)$. I have not thought about how often this happens. It is related to the relative norm being surprisingly mapped down to the base field.



Please check the above claim for this chosen $P$. If you think that the conjecture holds for this $P$ as well, let me know!
