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

Let $A$ be a $227\times 227$ matrix with entries in $\mathbb{Z}_{227}$ such that all of its eigen values are distinct. What would be its trace? I think it is zero by adding all 227 elements but i am not sure.

Edited: Here I have assumed that eigenvalues are in a base field.

-
What exactly do you mean by eigenvalues in $Z_{227}$? – copper.hat May 10 '12 at 21:51
they are distinct element of $\mathbb{Z}_{227}$ – srijan May 10 '12 at 21:55
And what would be the eigenvalues of $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ in this context? – copper.hat May 10 '12 at 21:58
You need to modify your question to state that not only are the eigenvalues distinct, but that there are 227 of them. – copper.hat May 10 '12 at 22:03
@srijan: Then please edit your question and mention that the eigenvalues are assumed to be in $\mathbb{Z}_{227}$. – Manos May 10 '12 at 22:03

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.

-
Nobody said the eigenvalues are in the base field. This answer is therefore wrong. – Marc van Leeuwen Jan 3 '15 at 10:30
@MarcvanLeeuwen If you read the comments to the OP, you can see the user said as much in the comments above. It appears that he has deleted his comment. – rschwieb Jan 3 '15 at 11:38
@rschwieb I have edited question statement. Thanks for pointing out. – srijan Jan 15 '15 at 9:32
@srijan thank you – rschwieb Jan 15 '15 at 11:05
@rschwieb You are welcome. :) – srijan Jan 15 '15 at 11:06

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 $\mathbb{Z}_{227}$.

-

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 $227$).

-
Why are all the diagonal elements distinct? – Manos May 10 '12 at 21:45
They don't need to be, as far as I can see. – copper.hat May 10 '12 at 21:46
So i don't see the validity of the above argument. As copper.hat noted earlier, the eigenvalues might not even be in $\mathbb{Z}_{227}$. – Manos May 10 '12 at 21:49
Assume that eigen values are from $\mathbb{Z}_{227}$. I got that question in our entrace exam for indian institute of techonology kharagpur india. – srijan May 10 '12 at 21:59

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

-