1
$\begingroup$

Let $\mathbf{A}$ be a hermitian positive semi-definite matrix and $\mathbf{B}$ be a hermitian positive definite matrix. Then I am interested in the eigenvalues of matrix $\mathbf{C}(t)=\mathbf{A}-t\mathbf{B}$ where $t$ is a real parameter. Note that when $t=0$, $\mathbf{C}$ is a positive semi-definite matrix. As $t$ increases, the eigenvalues of $\mathbf{C}$ moves towards the left on the real line and crosses the origin. At one point, the last positive eigenvalue of $\mathbf{C}$ will also cross origin at which point, $\mathbf{C}$ will become negative semi-definite. I want the smallest $t$ for which this happens. In other words, I want to solve \begin{align} \max_{t}~t \\ \lambda_{max}(\mathbf{C}(t))\geq 0 \end{align} where $\lambda_{max}(.)$ denotes the largest eigenvalue of the argument.

$\endgroup$
1
$\begingroup$

$\det(C(t))$ is a polynomial in $t$, of degree $n$ (where your matrices are $n \times n$). You want the greatest positive root of this polynomial.

You could also call this the greatest eigenvalue of $B^{-1/2} A B^{-1/2}$.

$\endgroup$
  • $\begingroup$ I got it as largest eigenvalue of $B^{-1}A$. How do I prove this? $\endgroup$ – dineshdileep Jul 19 '13 at 6:22
  • $\begingroup$ $M_1 M_2$ and $M_2 M_1$ always have the same eigenvalues (the same nonzero eigenvalues, if $M_1$ is $m \times n$ and $M_2$ is $n \times m$), but $B^{-1/2} A B^{-1/2}$ is Hermitian. $\endgroup$ – Robert Israel Jul 19 '13 at 14:58
  • $\begingroup$ still no success, how do you prove that it is the largest eigenvalue of $B^{-1/2}AB^{-1/2}$ $\endgroup$ – dineshdileep Jul 24 '13 at 8:29
  • $\begingroup$ $\det(A - \lambda B) = \det(B^{1/2}(B^{-1/2} A B^{-1/2} - \lambda I) B^{1/2}) = \det(B) \det(B^{-1/2} A B^{-1/2} - \lambda I)$ $\endgroup$ – Robert Israel Jul 25 '13 at 2:30
  • $\begingroup$ i am sorry if it so simple, but i am not getting it? $\endgroup$ – dineshdileep Aug 7 '13 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.