# Shifting Eigenvalues

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.

$\det(C(t))$ 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 $B^{-1/2} A B^{-1/2}$.
• I got it as largest eigenvalue of $B^{-1}A$. How do I prove this? – dineshdileep Jul 19 '13 at 6:22
• $M_1 M_2$ 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. – Robert Israel Jul 19 '13 at 14:58
• still no success, how do you prove that it is the largest eigenvalue of $B^{-1/2}AB^{-1/2}$ – dineshdileep Jul 24 '13 at 8:29
• $\det(A - \lambda B) = \det(B^{1/2}(B^{-1/2} A B^{-1/2} - \lambda I) B^{1/2}) = \det(B) \det(B^{-1/2} A B^{-1/2} - \lambda I)$ – Robert Israel Jul 25 '13 at 2:30