## Problem on Simultaneous Diagonalization

Let $m,n$ be two integer, find the number $f(m,n)$ suatisfying that: for any $a\leq f(m,n)$ and $m\times n$ complex matrices $A_1,...A_a$ satisfying $$\sum_{i=1}^a A_i^{+}A_i=I,$$ where $A^{+}$ denotes the conjugate transposed matrix of $A$, then there exists complex unitary $U=[u_{jk}]_{a\times a}$ such that $B_i^{+}B_i$ commute, which implies they are Simultaneous Diagonalized, where $B_i=\sum_{k=1}^a u_{ik}A_k$.

For instance, we ask three $3\times 3$ complex matrices $A_1.A_2,A_3$ with $A_1^{+}A_1+A_2^{+}A_2+A_3^{+}A_3=I$, whether there always exists complex unitary $U=[u_{jk}]_{3\times 3}$ such that $B_1^{+}B_1$ and $B_2^{+}B_2$ commute, where $B_i=\sum_{k=1}^a u_{ik}A_k$.

It is well known that $f(m,n)\geq 2$, if $f(m,n)\geq 3$ holds for all $m,n$? Especially, I wanna know if $f(3,3)\geq 3$ and if $f(3,4)\geq 3$.

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Can somebody translate from Sindarin into English? – Wadim Zudilin Oct 14 2010 at 14:03
Since symmetric commuting matrices are indeed simultaneously diagonalized by a unitary matrix, I understand that $A^+$ denotes the transpose of $A,$ and that "are communicated" is an erratum to be read commute. It would be nice if user Gondolf confirmed. The question seems appropriate, and I don't really see strong reasons for closing. Please, gentlemen, be patient. – Pietro Majer Oct 14 2010 at 19:54
You are right, thank you:-) – gondolf Oct 15 2010 at 0:43