trace

**alexandrabel90** · December 1st 2010, 08:39 AM

let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the coeff of x^(n-1) in the characteristic polynomial is -tr(A)

**FernandoRevilla** · December 1st 2010, 09:14 AM

Originally Posted by alexandrabel90

let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the coeff of x^(n-1) in the characteristic polynomial is -tr(A)

We have

$\begin{vmatrix} x-a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} &x-a_{22} & \ldots & a_{2n} \\ \vdots&&&\vdots \\ a_{n1} & a_{n2} &\ldots & x-a_{nn}\end{vmatrix}=\\<br /> (x-a_{11})(x-a_{22})\cdot\ldots\cdot(x- a_{nn})+\ldots$

and the terms we haven't written have degree $\leq n-2$ .

Regards.

Fernando Revilla

**Drexel28** · December 1st 2010, 02:27 PM

Originally Posted by alexandrabel90

let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the coeff of x^(n-1) in the characteristic polynomial is -tr(A)

Another method (the first is similar to that of Dr. Revilla's) is that you can first recall that $p_A(x)$ is monic and it's roots are precisely the mulitiset spectrum $\sigma\left(A\right)$ of $A$ . and so

$\displaystyle p_A\left(x\right)=\prod_{\lambda\in\sigma\left(A\r ight)}(x-\lambda)$

We can then recall Vieta's formula so that (remembering that $p_A(x)$ is monic) that $-\text{coeff}(x^{n-1})=\Pi_1(\lambda_1,\cdots,\lambda_n)$ where $\Pi_1(\lambda_1,\cdots,\lambda_n)$ is the first symmetric polynomial. Thus,

$\displaystyle -\text{coeff}(x^{n-1})=\Pi_1(\lambda_1,\cdots,\lambda_n)=\sum_{\lambd a\in\sigma(A)}\lambda=\text{tr}(A)$

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