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Let $f:M_n(\mathbb{C}) \rightarrow \mathbb{C}$ be a function such that $f(X+Y)=f(X)+f(Y)$ for all matrices $X, Y$ in $M_n(\mathbb{C})$. Find $f$.

Rengo
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  • Probably the trace – Luukdr Feb 13 '23 at 18:48
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Feb 13 '23 at 18:50
  • @Luukdr yeah I thought if that as well, but can't seem to find a proof. – Rengo Feb 13 '23 at 19:06
  • The trace only looks at the diagonal elements of a matrix. So $\mathrm{Tr}(A+B) = \sum_{i=1}^n (A+B){ii} = \sum{i=1}^n A_{ii}+B_{ii} = \sum_{i=1}^n A_{ii} +\sum_{i=1}^n B_{ii} = \mathrm{Tr}(A)+\mathrm{Tr}(B)$, where $A_{ii}$ denotes the $i$th diagonal entry. – Luukdr Feb 13 '23 at 19:29
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    Just to clarify, do you need one $f$ or all $f$ satisfying that condition? ("All" might be hard without an additional condition) – Bruno B Feb 13 '23 at 20:07
  • How about the extended form $f(aX+bY)=af(X)+bf(Y)$ for all $a$ and $b$ complex numbers and $X$ and $Y$ complex matrices? – Rengo Feb 13 '23 at 20:19
  • Then it's still true for $f = \mathrm{Tr}$. – Luukdr Feb 13 '23 at 20:51
  • Yes but I need to find all f, not just one. – Rengo Feb 14 '23 at 11:42

1 Answers1

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The problem is that (at least under the axiom of choice) this equation admits solutions that will be way too "not nice" so to speak, as this equation is too generic.

The same equation from $\mathbb{R}$ to $\mathbb{R}$ has been studied extensively, and there are non-linear highly irregular solutions (we're talking "dense graphs in $\mathbb{R}^2$" irregular) under AC by using Hamel bases, and the generalisation to complex numbers as well as higher dimensions is quite easy, by slapping an irregular solution $f_0$ of the Cauchy equation onto each vector of a basis of your space and extending by additivity to the whole matrix space, so something like $$f(A) := \sum_{k,l} \left(f_0\left(\operatorname{Re}(a_{k,l})\right) + i f_0\left(\operatorname{Im}(a_{k,l}\right)\right) E_{k,l}$$

However, just like what is presented in that post I linked above, there are a few regularity assumptions that'll guarantee the solution to be linear, most prominently continuity but not only that.
Assuming you have one of these conditions, the solutions are exactly all the linear functionals, which is more than good enough to describe the solution set, but if you still want a simple "characterisation", then this post for example shows that you can describe any and every linear functional on a square matrix space as $f : A \mapsto \operatorname{Tr}(BA)$ for a certain matrix $B$.

Bruno B
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