Given a positive definite matrix $V(x)=x^TBx>0$ for any vector $x≠0$. we have $λ_{min}||x||^2≤V(x)≤λ_{max}||x||^2$ for any $x≠0$ , Then we could derive a upperbound
$$||x||\leq\ \bigg(\frac{V(x)}{\lambda_{min}}\bigg)^{\large\frac{1}{2}}$$
My question is: isn't the form $x^TBx=\lambda ||x||^2$ only works for the eigenvectors? Which means there are some eigenvalues $\lambda$ associate with. What if the $x$ is not the eigenvector, is above upperbound still hold for all $x$ satisfy $x^TBx>0$?
Best Answer
If $B \in \mathbb{R}^{n \times n}$ is symmetric, then $\mathbb{R}^n$ has an orthonormal basis $\{q_1, \dots, q_n\}$ of eigenvectors of $B$. Thus if $\{\lambda_1, \dots, \lambda_n\}$ are the corresponding eigenvalues and $x \in \mathbb{R}^n$, then $$x = \sum_{j = 1}^{n}(x, q_j)q_j,$$ $$Bx = \sum_{j = 1}^{n}(x, q_j)\lambda_jq_j,$$ $$V(x) = (Bx, x) = \sum_{j = 1}^{n}(x, q_j)^2\lambda_j.$$ The bounds on $V(x)$ follow immediately.
If $B$ is not symmetric, then set $S = \frac{1}{2}(B + B^T)$, the symmetric part of $B$. Since $B - S = \frac{1}{2}(B - B^T)$ is antisymmetric, $((B - S)x, x) = -(x, (B - S)x) = -((B - S)x, x)$ for all $x \in \mathbb{R}^n$. Thus $((B - S)x, x) = 0$ for all $x \in \mathbb{R}^n$. Thus $(Bx, x) = (Sx, x)$ for all $x \in \mathbb{R}^n$. Now the bounds for $V(x)$ in the symmetric case apply, though with eigenvalues of $S$ instead of $B$.