Inner product inequality

Question

Is there any inequality for $x^TAy$ where $x$ and $y$ are vectors and $A$ is positive definite matrix.

For example: $x^TAy\ge k||x||||y||$ where $k$ is a coefficient of (min or max) eigenvalue of $A$.

Answer 1

The thing with the biggest (by absolute value) eigenvalue definitely works (but with $\leq$ sign, of course). You can also write a similar inequality in general for singular value decomposition.

The theorem you are looking for is the bounding of the induced matrix norm by the eigenvalues:

http://en.wikipedia.org/wiki/Matrix_norm#Induced_norm

The derivation would go as follows:

Cauchy-Schwarz: $$x(Ay)\leq |x| |Ay|$$ Matrix norm definition ($|A|=\sup(|Ax|/|x|)$): $$|Ay|\leq |A||x|$$ Equality: $$|A|=\sigma_{max}(A)$$

Therefore $$xAy\leq \sigma_{max}(A)|x||y|$$ This is true for any $A$, not just for square matrices.

Similarly, you get $$xA^{-1}y\leq \sigma_{min}^{-1}(A)|x||y|$$ for an invertible matrix.