Find out trace of a given matrix $A$ with entries from $\mathbb{Z}_{227}$

Solution 1:

The trace of a matrix is defined to be the sum of its diagonal elements. The trace is equal to the sum of the eigenvalues, i.e. the roots of the characteristic polynomial, when the characteristic polynomial splits over the underlying field $F$. Consequently, we can not say anything in the present case without further information.

For example, if all eigenvalues are in $\mathbb{Z}_{227}$ and they are distinct, then yes, the trace is equal to zero. But in general, the eigenvalues will not be all of them inside $\mathbb{Z}_{227}$.

Solution 2:

Try the matrix with 113 'block elements' of the form $\begin{bmatrix} 0 & n \\ n & 0 \end{bmatrix}$, with $n$ running from $1$ to $113$, and the last diagonal element $114$. Then the trace is 114 and all eigenvalues are distinct.

The eigenvalues are $\{\pm n i\}_{n=1}^{113} \cup \{114\}$.

Solution 3:

Since your field has exactly $227$ elements and all $227$ eigenvalues are distinct, the trace of $A$ is exactly the sum of all $227$ elements in $\mathbb{Z}_{227}$. It well known that the sum of the first $n$ positive integers is $n(n+1)/2$, so here $$ 0+1+2+\cdots+226=\frac{226\times227}2. $$ Since we are doing arithmetic modulo $227$, this sum is zero (being a multiple of $227$).

Solution 4:

Yes: all eigenvalues distinct means that every value from $\mathbb{Z}_{227}$ appears exactly once, so you just need to compute $\Sigma_{i=0}^{226} i=226(227)/2=0 \pmod{ 227}$.

Addendum: at one point the user made a comment about adding the hypothesis that the eigenvalues are in the field, but now that comment appears to be gone. Without the change, this line of reasoning is not useful.