# Rayleight Ritz Ratio and smallest eigenvalue for a set of given matrices

I am familiar with Rayleigh Ritz Ratio for hermitian matrices. Let $A_1$ be a given $N \times N$ hermitian matrix. Then the smallest eigenvalue of $A_1$ is given by \begin{align} \lambda_{min}(A_1)=\min_{x^Hx=1}x^HA_1x \end{align} This led me to the following question. Let me define a new quantity \begin{align} \lambda_{min}(A_1,A_2)=\min_{x^Hx=1}\max(x^HA_1x,x^HA_2x) \end{align} where $A_2$ is another given hermitian matrix. Is this quantity known before?. Is any there intuitive way of looking at it?

I was able to figure out that I could solve it using semi-definite optimization. Any comments on what happens when I try it for $k$ matrices $A_1,A_2,\ldots,A_k$

Here is a partial answer. But let me first modify the question by replacing the reals by the complex. Hence, $x$ runs over ${\mathbb C}^n$ and $A_1,A_2$ are Hermitian. Then $A_1,A_2$ can be viewed as the "real" and "imaginary" parts of a single matrix $M=A_1+iA_2$, that is $$A_1=\frac12(M+M^*),\qquad A_2=\frac1{2i}(M-M^*).$$ The advantage of working within the complex numbers is that we know (Toeplitz-Hausdorff Theorem) that the image ${\cal H}(M)$ of the unit sphere of ${\mathbb C}^n$ under the numerical map $x\mapsto x^*Mx=x^*A_1x+ix^*A_2x$ is a convex compact subset of the complex plane.

Now, $\lambda_\min(A_1,A_2)$ (under the complex definition) is the minimum of the convex function $f(z)=\max(\Re z,\Im z)$ over ${\cal H}(M)$. We easily see that it is achieved on the boundary of ${\cal H}(M)$, at a point which is either the downmost point, or the leftmost point, or the most SW point along the diagonal $\Delta$ defined by $\Re z=\Im z$.

Specifically, we have the following alternative, where $x$ (resp. $y$) denotes some unit eigenvector of $A_1$ (resp. $A_2$) associated with $\lambda_\min(A_1)$ (resp. $\lambda_\min(A_2)$) :

• either $x^*A_1x>x^*A_2x$, and then $\lambda_\min(A_1,A_2)= \lambda_\min(A_1)$,
• or $y^*A_2y>y^*A_1y$, and then $\lambda_\min(A_1,A_2)= \lambda_\min(A_2)$,
• or $x^*A_1x\le x^*A_2x$ and $y^*A_2y\le y^*A_1y$, and then $t:=\lambda_\min(A_1,A_2)$ is the number such that $(t,t)$ is the down-left point of the interval ${\cal H}(M)\cap\Delta$.
• As a complement to Denis's approach, I think that joint numerical ranges could help to extend his alternative to $k$ matrices. See this paper, for example: sciencedirect.com/science/article/pii/S0024379503006797 – Felix Goldberg Dec 5 '12 at 11:55
• @Denis Serre Excellent answer!! Thanks for that one. – dineshdileep Dec 6 '12 at 12:05
• @Felix Goldberg, I will take a look at it. – dineshdileep Dec 6 '12 at 12:05
• @Denis Serre Actually I had one more doubt, the image is closed convex compact set, does it mean, every point in it has a corresponding vector in the unit sphere which gives that point? – dineshdileep Dec 6 '12 at 12:07
• @dineshdileep. Yes indeed. This is a bit surprising, because ${\cal H}(M)$ is the image of a non-convex set (the unit sphere) by a nonlinear map (the numerical map). But this is the contents of the TH Theorem. Actually, this is a deep result, with many interesting consequences and relations. For an extension of this result, see my recent paper with Th. Gallay in CPAM 65 (2012), pp 287-336. – Denis Serre Dec 6 '12 at 14:05