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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

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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.

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    What are the **values** of a characteristic polynomial? The eigenvalues? i.e. the zeros of the characteristic polynomial? What do B and E stand for in your table? – 2017-01-10
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    I am trying to guess what might be happening. How often products of two eigenvalues of $P_5$ belong to the base field could heavily depend on how the characteristic polynomial of $P_5$ factors over the base field. If you have a lot of quadratic factors, then the product of the roots of such a factor will automatically be in the base field. But if the characteristic polynomial has an irreducible quartic factor then it should not happen. Have you tested with a $P_5$ whose characteristi polynomial has an irreducible quartic as a factor? – 2017-01-10
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    Yes, values for me is the eigenvalues. B means Base Field, E means extension field. I know that when I do the product of eigenvalues in the Base (or Ground) Field the result is in the Ground Field and when I multiply values from the Ground Field with values in the extension Filed the result is on the extension field. But, when I multiply values in the extension field with values in the extension field the result can be in the ground or in the extension field. I noticed that when those values are the eigenvalues of a characteristic polynomial, half of then will lie in the ground. – 2017-01-10
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    Note the P25 is the Kronecker product of P5 by itself. This way, I know that eigenvalues of P25 are the product of the eigenvalues of P5. – 2017-01-10
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    If when factoring characteristic polynomial there appear an irreducible polynomial then its roots lies in an extension field. I can determine which extension field is by discovering to what expoent this polynomial belongs to. – 2017-01-10

1 Answers 1

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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!