# Trace of a $227\times 227$ matrix over $\mathbb{Z}_{227}$ [duplicate]

well, I know that the trace is the negative of coefficient of $x^{226}$ of the characteristic polynomial the matrix, but I dont know how the Char.Poly looks like in this case.please give me some hint. Do I have to work in the splitting field of the characteristic polynomial and add the eigen values to get the trace?but I dont know how.

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## marked as duplicate by Marc van Leeuwen linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 3 '15 at 10:33

Let's do something smaller. Use 3 instead of 227. The matrix $$\pmatrix{0&0&0\cr0&1&0\cr0&0&2\cr}$$ has distinct eigenvalues and trace zero. The matrix $$\pmatrix{1&0&0\cr0&0&1\cr0&2&0\cr}$$ has distinct eigenvalues $1,i,-i$, where $i$ is a square root of minus one in an extension field, and it has trace 1. So if the eigenvalues are allowed to be in an extension field, the answer is not determined by the information given.
so, is there any specific answer of the trace?I think if the eigen values are in the field and distinct, then trace is $0$? – La Belle Noiseuse Jul 22 '12 at 11:13