# Minimum eigenvalue of a Affine Combination of two Hermitian matrices

Consider two $N \times N$ hermitian indefinite matrices $A_1$ and $A_2$. Consider their affine combination \begin{align} M(t)=(1-t)A_1+tA_2 \end{align} I am interested in the minimum eigenvalue of $M(t)$. I can write this as \begin{align} \lambda(t)=\min_{x ~\in~\mathbb{C}^{N\times 1}}~&x^HM(t)x \\\ &x^{H}x = 1 \end{align}

When I simulated this, I made the following observations

• It is a concave function of $t$
• It is a piece-wise linear function of $t$ (at least, that is what the plot looks like).

My question, 1) how do I explain this behavior 2) Any comments on $t$ where the maxima occurs?

Some Thoughts:

I have a feeling that this is connected to the Toeplitz-Hausdorff theorem. From it, it follows that the subset of 2-D plane \begin{align} \mathbb{S}=\{[x_1,x_2]\in \mathbb{R}^2\mid ~x_1=x^HA_1x,~x_2=x^HA_2x,~x^{H}x=1\} \end{align} is a closed compact convex subset of $\mathbb{R}^2$. Thus, it follows that minimum eigenvalue of $M(t)$ is \begin{align} \lambda(t)=(1-t)x_1+tx_2, ~~[x_1,x_2]\in \mathbb{S} \end{align} Now from here (if this is correct?), How do I conclude the observations, I made earlier?

-

This is related to so-called hyperbolic polynomials, studied by L. Gaarding in the fifties. More generally, let $\lambda(\xi)$ be the least eigenvalue of $A(\xi)=\sum_\alpha\xi_\alpha A^\alpha$, where $A^\alpha$ are Hermitian matrices and $\xi$ is a real vector. Then $\lambda$ is a concave function. It is generically strictly concave, except in the radial directions of course because of homogeneity $\lambda(s\xi)=s\lambda(\xi)$ for $s>0$. The strict concavity is related to the lack of commutativity of pairs $(A^\alpha,A^\beta)$. For instance, the Pauli matrices yield $\lambda(\xi)=-|\xi|$, which is clearly strictly concave away from rays ${\mathbb R}^+\xi$. On the opposite, if $[A^\alpha,A^\beta]=0$ for every pair, then $\lambda$ is piecewise linear.

Strict concavity occurs for instance when the least eigenvalue is simple for every $\xi\ne0$, or if it has a constant multiplicity. Then $\lambda$ is analytic away of the origin, with a Hessian matrix of rank $n-1$. It turns out that this property implies a so-called Strichartz inequality for the solutions of the system of Partial Differential Equations $$\frac{\partial u}{\partial t}+\sum_\alpha A^\alpha\frac{\partial u}{\partial x_\alpha}=0.$$ Obviously, the symbol of the system is $\det(\tau I_n+A(\xi))$.Thus the characteristic manifold is related to the eigenvalues of $A(\xi)$, in particular to $\lambda(\xi)$.

The other eigenvalues satisfy more involved inequalities such as Weyl, Lidskii, Ky Fan -type inequalities. For instance, the sum of the $k$ least eigenvalues is concave too. This is a part of Alfred Horn's conjecture, now a theorem thanks to the work of many people, including Knutson & Tao.

The last part of the question, that about Toeplitz-Hausdorff theorem, is unclear. ${\mathbb S}$ is not a singleton, so what means \begin{align} \lambda(t)=(1-t)x_1+tx_2, ~~[x_1,x_2]\in \mathbb{S} \qquad ? \end{align} The equality certainly holds true for teh particular point $(x_1,x_2)$ obtained by taking $x$ a unit eigenvector associated with $\lambda(t)$, but what else ? To see some deep relations between Toeplitz-Hausdorff and hyperbolic polynomials, have a look to our paper with Th. Gallay, Numerical measure of a complex matrix, in Comm. Pure Appl. Math. 65 (2012), no. 3, 287–336.

-
Can we relate it to Toeplitz-Hausdorff theorem and comment about the observations? –  dineshdileep Dec 19 '12 at 7:01
great answer -- minor typo (teh -> the) in last paragraph –  J.J. Green Dec 19 '12 at 11:39
@Denis Thanks, I got my mistake. –  dineshdileep Dec 21 '12 at 5:21

The function $t\mapsto x^TM(t)x$ is affine for every $x$. Inf of affine functions is usually concave. (I mean it is concave under some broad assumptions which are certainly satisfied here).

About "piecewise-linear", it is not true, even for simple 2 times 2 matrices.

-