How can a show the following assertion?

If $(a_{ij})$ is a positive definite matrix and $\lambda,\Lambda$, the minimum and maximum eigenvalues, then $$\lambda|x|^2\leq\sum_{i,j}^na_{ij}x^ix^j\leq\Lambda|x|^2 $$ In other words $$\lambda\langle x,x\rangle\leq\langle x,Ax\rangle\leq\Lambda\langle x,x\rangle$$ where, $x=(x^1,x^2,\ldots,x^n)$


  • $\begingroup$ There may not be any eigenvalues. $\endgroup$ – Jeff Jun 28 '13 at 18:49
  • $\begingroup$ even if $A$ is normal, then you need to consider complex eigenvalues, which OP doesn't seem to do. $\endgroup$ – justt Jun 28 '13 at 18:54
  • $\begingroup$ Just diagonalize $A$. But in an orthonormal basis. $\endgroup$ – Julien Jun 28 '13 at 19:45

$A$ is diagonalized in an orthonormal basis : so we have an orthogonal matrix $P$ such that $$A = P^T DP.$$ With $\left( \begin{matrix} \lambda & & (0) \\ & \ddots & \\ (0) & & \Lambda \end{matrix} \right)$ (the eigenvalues are ordered).

So $\langle x, Ax\rangle =x^TAx = x^TP^TDPx = (Px)^T D\,(Px) $.

But for every $y$, we have $\lambda ||y|| \leq y^TDy \leq \Lambda ||y||$ (this is easy to write).

So if we take $y=Px$ and notice that $||x|| = ||Px||$, we are good.


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