Let $\displaystyle A \in M_{n \times n}(\mathbb{R})$. Show that if A has n eigenvalues $\displaystyle \lambda_1,\lambda_2,...,\lambda_n$ in $\displaystyle \mathbb{R}$ then $\displaystyle \lambda_1+\lambda_2+...+\lambda_n = traceA$ and $\displaystyle \lambda_1 \lambda_2 ... \lambda_n = detA$.

Not really sure where i'm going with this question. I am thinking of using the fact that for the polynomial equation obtained from $\displaystyle det(\lambda I_n - A)=0$ has a coefficient of trace(A) for the $\displaystyle \lambda^{n-1}$ term, but not really sure how this is useful.