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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.

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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

4 Answers 4

up vote 2 down vote accepted

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

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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 $227$).

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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

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

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

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