Let $Q:\ \mathbb R^n\longrightarrow\mathbb R$ is a positive definite quadratic form and $\mathcal B$ is a basis of $\mathbb R^n$. Then how to show that all coefficients of the characteristic polynomial $\,p(t)=\big|\big[Q\big]_{\mathcal B}-t I_n \big|\, $ are non-zero and alternating in sign ?
My attempt :
Earlier, I showed that every eigenvalue of $ \big[Q\big]_{\mathcal B}$ is positive. Then $$p(t)\,=\,(-1)^n(t-\lambda_1)\cdots(t-\lambda_n) $$ where $\lambda_i>0$ for all $i$.
When $n=1$, the statement is clearly true. Suppose the statemnent is true for $n$. Consider \begin{align} p(t)\,&=\,(-1)^{n+1}(t-\lambda_1)\cdots(t-\lambda_n)(t-\lambda_{n+1}) \\ &=(-1)^{n+1}\Big[(t-\lambda_1)\cdots(t-\lambda_n)t-(t-\lambda_1)\cdots(t-\lambda_n)\lambda_{n+1}\Big]. \end{align} Suppose \begin{align} (t-\lambda_1)\cdots(t-\lambda_n)\,=\,a_0+a_1t+\cdots+a_nt^n \end{align} then we have \begin{align} p(t)\,=\,(-1)^{n+1}\Big[-\lambda_{n+1}a_0+(a_0-\lambda_{n+1}a_1)t+\cdots+ (a_{n-1}-\lambda_{n+1}a_n) t^n+a_nt^{n+1} \Big]. \end{align} Assuming without loss of generality that $a_0<0$. By inducing hypothesis, $a_1>0,a_2<0,\cdots$. Hence \begin{align} -\lambda_{n+1}a_0\,&>\,0 \\ a_0-\lambda_{n+1}a_1\,&<\,0 \\ &\vdots \end{align} Therefore the statement is true for $n+1$. So it is true for every $n$.
Is my proof ok ? Is there others way to solve this problem ? Pleas share your opinion. Thanks.