If $A$ is an $n$-by-$n$ matrix with characteristic polynomial $(-1)^nt^n+b_{n-1}t^{n-1}+\cdots+b_1t+b_0$, show that $tr(A)=(-1)^{n-1}b_{n-1}$

Asked Apr 22 '20 at 19:05

Active Apr 22 '20 at 19:19

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Let $A$ be an $nxn$ matrix with characteristic polynomial $$f(t)=(-1)^nt^n+b_{n-1}t^{n-1}+\cdots+b_1t+b_0$$

Show that $tr(A)=(-1)^{n-1}b_{n-1}$

I think I should be able to get this from the characteristic polynomial given however I don't know where to start.

My thinking is to start by trying to replace $f(t)$ with $det(A-\lambda I)$ and attempting to work backwards to get to A, and then solve for trace. My other thought is to use the Pattern Method to try and get $f(t)$

edited Apr 22 '20 at 19:19

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asked Apr 22 '20 at 19:05

Thomas Nowak

If $A$ is an $n$-by-$n$ matrix with characteristic polynomial $(-1)^nt^n+b_{n-1}t^{n-1}+\cdots+b_1t+b_0$, show that $tr(A)=(-1)^{n-1}b_{n-1}$

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