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The matrices I discuss are all $N\times N$ hermitian matrices. Consider two (hermitian) matrices $A_1$ and $A_2$. For a real scalar $t$, define the following function for the matrix $A_1+t*A_2$ \begin{align} \lambda(t)=\min_{u^Hu=1}~u^H(A_1+t*A_2)u \end{align} For a given $t$, this is essentially the smallest eigenvalue of $A_1+t*A_2$. This should be a concave function. Now define the function \begin{align} f(t)=||A_1+t*A_2||_F^2 \end{align} I randomly generated two hermitian matrices in my simulation software (matlab), and plotted this for a range of $t$ The image. You can see something happening here. $f(t)$ (blue curve) appears to be a convex function. The main thing being the value of $t$ at which $f(t)$ attains its minimum and $\lambda(t)$ (red curve) attains its maximum seems to be same. Is there any explanation for this. I tried it several times (more than 15 times at least), the nature of the resulting graph is always same. Can anyone give a possible explanation for this.

UPDATE--- I checked the exact values. They are close. But not the same. But still any possible explanation for this behaviour.

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Note that you can get proper spacing for the norm bars using \lVert and \rVert. – joriki Dec 2 '12 at 14:49
up vote 1 down vote accepted

As long as you don't quantify "close", there's nothing to explain here. It's clear that both graphs are asymptotically linear in both directions and must have a cross-over region in between. Thus if you zoom out far enough that both cross-over regions look small, their graphs will inevitably look like this.

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This came from a random gaussian matrix. As soon as changed the variances, the behavior started to change. So I guess there is not much of thing inside it. Thanks for the answer – dineshdileep Dec 5 '12 at 4:08

I think you plotted $g: t \mapsto \Vert A_1 + t A_2 \Vert_F$ as the upper curve, not its square $f$.

For the special case $A_2 = I$ (identity matrix), you will not get this behavior. In that case $\lambda(t) = t + \mu_{min}(A_1)$, where $\mu_{min}(A)$ is the smallest eigenvalue of the matrix $A$. More generally, the slope of $\lambda$ will be near $\mu_{max}(A_2)$ for large negative $t$ and near $\mu_{min}(A_2)$ for large positive $t$. If $A_2$ is positive (negative) definite, then $\lambda$ can be expected to increase (decrease) monotonically.

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Thanks for pointing out that, I was not plotting the square. $A_1$ and $A_2$ are hermitian indefinite. But I observed the behaviour you pointed when $A_2$ is positive definite (or negative). I guess that can proved from interlacing theorem. – dineshdileep Dec 4 '12 at 16:31

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