# Lower bound on matrix and vector product for vector on unit sphere

Suppose, that $A$ is a positive definite matrix, and $x$ is a vector which belongs to unit sphere ($||x|| = 1$). Norm is Euclidean. Does it exist a nonzero lower bound for $x^T A x$ product? I.e. does it exist a number $c > 0$ which depends on $A$ such that $c < x^T A x$ for all $x$ with Euclidean norm one.

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Yes, and this real number $c$ is any number such that $c<\lambda_{\min}(A)$, where $\lambda_{\min}(A)$ is the smallest eigenvalue of $A$. To see that, write $A=P^tDP$, where $P$ is orthogonal and $D$ diagonal. As $P$ is orthogonal, $\lVert Px\rVert=\lVert x\rVert$ so we want $c$ such that $c< u^tDu$ for all $u$ of Euclidian norm $1$. As $u^tDu\geq \lVert u\rVert^2\lambda_{\min}(A)=\lambda_{\min}(A)$, we have the wanted result.
I use the definition of "positive definite" that include symmetric real matrices, or Hermitian complex ones. So $A$ is diagonalizable and we may assume that $x^TAx=\sum \lambda_j x_j^2$, where $\lambda_j>0$. From this the statement follows.