We have $N$ Hermite matrices $A_i$ and $N$ weight values $w_i$, $1\leq i\leq N$, $\sum_{i=1}^N w_i=1$.

Then we can obtain a new Hermite matrices $\sum_{i=1}^N w_iA_i$. let us assume $\lambda$ is the minmum non-zero eigenvalue of $\sum_{i=1}^N w_iA_i$, and vector $X$ is the corresponding eigenvector.

My question is how to select $w_i$， so that $\max_iX^HA_iX$-$\min_iX^HA_iX$ is as minimal as possible.

thanks for your answer.

We have $N$ Hermitian matrices $A_i$ and $N$ weight values $w_i$, $1\leq i\leq N$, $\sum_{i=1}^N w_i=1$.

Then we can obtain a new Hermite matrices $\sum_{i=1}^N w_iA_i$. let us assume $\lambda$ is the minmum non-zero eigenvalue of $\sum_{i=1}^N w_iA_i$, and vector $X$ is the corresponding eigenvector.

My question is how to select $w_i$， so that $\max_iX^HA_iX$-$\min_iX^HA_iX$ is as minimal as possible.

thanks for your answer.

We have $N$ Hermite matrices $A_i$ and $N$ weight values $w_i$, $1\leq i\geq N$,  $\sum_{i=1}^N w_i=1$.

Then we can obtain a new Hermite matrices $\sum_{i=1}^N w_iA_i$. let us assume $\lambda$ is the minmum non-zero eigenvalue of $\sum_{i=1}^N w_iA_i$, and vector $X$ is the corresponding eigenvector.

My question is how to select $w_i$， so that $\max_iX^HA_iX$-$\min_iX^HA_iX$ is as minimal as possible.

thanks for your answer.

We have $N$ Hermite matrices $A_i$ and $N$ weight values $w_i$, $1\leq i\leq N$,   $\sum_{i=1}^N w_i=1$.

Then we can obtain a new Hermite matrices $\sum_{i=1}^N w_iA_i$. let us assume $\lambda$ is the minmum non-zero eigenvalue of $\sum_{i=1}^N w_iA_i$, and vector $X$ is the corresponding eigenvector.

My question is how to select $w_i$， so that $\max_iX^HA_iX$-$\min_iX^HA_iX$ is as minimal as possible.

thanks for your answer.