2

Suppose I have a positive definite matrix $A$, so $$x^TAx \geq C_A|x|^2$$ holds for all $x$. $A^{-1}$ is positive definite too: $$y^TA^{-1}y \geq C_B|y|^2,$$ is there any way I find the constant $C_B$ in terms of $C_A$?

studentX
  • 159
  • Are you assuming $A$ is Hermitian (i.e. symmetric for real $A$)? If so, then $A^{-1}$ is guaranteed to be positive definite if $A$ is. Also it may be easier to figure out the constants – user2566092 Aug 30 '13 at 16:21
  • Yes it is symmetric and real. I want to find the constant $C_B$ – studentX Aug 30 '13 at 16:21
  • 1
    Well, the largest value for $C_A$ is the smallest eigenvalue of $A$ and the largest value for $C_B$ is the reciprocal of the largest eigenvalue of $A$, so, in general, you cannot write $C_B$ in terms of $C_A$. (Unless you have some special relationship between these eigenvalues.) – copper.hat Aug 30 '13 at 16:25
  • You can't. Just consider a 2x2 diagonal matrix with entries $C_A$ and $C_B^{-1}$ that satisfies $C_B^{-1} \ge C_A > 0$. Aside from this inequality, there is no more information about $C_B$ you can extract from $C_A$. – achille hui Aug 30 '13 at 16:27
  • Thank you for the comments. – studentX Aug 31 '13 at 10:46

1 Answers1

3

All eigenvalues are positive; in general $C_A$ is the smallest eigenvalue of $A$. So $C_B$ is $1/\lambda_1$ where $\lambda_1$ is the largest eigenvalue of $A$. In general this can't be written in terms of $C_A$ since the largest and smallest eigenvalues in general don't have to be related in any way.

user2566092
  • 26,142