Find all complex matrices $A$ such that $n\operatorname{Tr}(AB) = \operatorname{Tr}(A)\operatorname{Tr}(B)$ for all $B$. Consider a bilinear form $f(A,B) = n\operatorname{Tr}(AB) - \operatorname{Tr}(A)\operatorname{Tr}(B)$ defined on $M_n(\mathbb{C})$.
I need to find the set $U^\perp$ of all matrices $A$ such that $f(A,B) = 0$ for every $B \in M_n(\mathbb{C})$, or, more specifically, find $\dim(U^\perp)$.
Other than the zero matrix I can't find any other general matrix. So $\dim (U^\perp) = 0$ I'd assume?
 A: I know that this is no complete answer, but just a hint that you might not have found all matrices such that $f(A,B) = 0 ~ \forall B$:
Take $c \in \mathbb{C}$ to be any complex number and let $I$ be the $n \times n$ identity matrix. Then
$$Tr(c \cdot I) = n \cdot c$$ 
and
$$Tr(c \cdot I \cdot B) = Tr(c \cdot B) = c \cdot Tr(B)$$
Hence,
$$f(c \cdot I, B) = n \cdot c \cdot Tr(B) - n \cdot c \cdot Tr(B) = 0$$
A: As in GenericNickname's answer, the matrices $cI$ satisfy the property you want.
Let's show that those are the only ones that satisfy this property: consider the matrix $B=E_{ij}$, which has $1$ in the $(i,j)$ entry, and $0$ everywhere else. By choosing $i\neq j$, we get that $$n{\rm Tr}(AE_{ij})={\rm Tr}(A)\cdot{\rm Tr}(E_{ij})={\rm Tr}(A)\cdot 0=0.$$ But, the only nonzero entry in the diagonal of $AE_{ij}$ is $a_{ji}$, therefore $a_{ji}=0$ for $j\neq i$.
Now, if $i=j$, then $$na_{ii}=n{\rm Tr}(AE_{ii})={\rm Tr}(A)\cdot{\rm Tr}(E_{ii})={\rm Tr}(A).$$ Since this holds for all $i$, this shows that all the $a_{ii}$ should be equal to each other. So $A$ is a multiple of the identity matrix.
Therefore ${\rm dim}(U^{\perp})=1$.
