I'm stuck with how to show that the trace of a matrix $A$ is the coefficient of $x^{n-1}$ in the characteristic polynomial of $A$.
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Do you already know about the Jordan canonical form? – 5xum Sep 07 '15 at 08:33
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Yes. Does it have something to do with similarity?? – Sep 07 '15 at 08:34
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2This could be trivial if you define the trace as the sum of the eigenvlaues – user217285 Sep 07 '15 at 08:35
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This is quite easy to prove using Leibniz' formula for the determinant. – Crostul Sep 07 '15 at 08:36
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Why is the trace the sum of the eigenvalues? And why is this the coefficient of the (n-1) term? – Sep 07 '15 at 08:39
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See also: http://math.stackexchange.com/questions/1616930/trace-and-the-coe%EF%AC%83cients-of-the-characteristic-polynomial-of-a-matrix – Martin Sleziak Jan 18 '16 at 15:51
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Duplicate of http://math.stackexchange.com/questions/1426006/tracea-coefficient-of-xn-1-in-charpolya – user26857 Jun 25 '16 at 20:01
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Hint:
Consider $\;\chi_A(x)=\det(xI-A)$. The $x$s appear only on the diagonal. Now the determinant is a signed sum of products of coefficients, each one in a different column and a different row, in all possible ways. As there are $n$ rows and columns, to get $x^{n-1}$ in such a product, we have to take $n-1$ coefficients on the diagonal, hence take the $n$ coefficients $x-a_{ii}$ and expand: $$(x-a_{11})\dotsm(x-a_{nn}).$$ Now this comes down to one of Vieta's relations between roots and coefficients of a polynomial.
Bernard
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