About inner product I'm stuck in the next problem,
Let $n$ a positive integer and $V=M_n(\mathbb{C})$. We define the inner product by $\langle A,B\rangle=\operatorname{tr}(A^t\overline{B})$. Let $W$ the subspace of $V$ such that if $A\in W$, then $\operatorname{tr}(A)=0$.
Find $W^\perp$.
Can you help me?
Greetings.
 A: Note that the dimension of $W$ is $n^2-1$ (try to find a basis). Note that $\text{dim}(W)+\text{dim}(W^{⊥})=\text{dim}(V)$, by the dimension formula. So $\text{dim}(W^{⊥})=1$. It is then not difficult to show that for any $A \in W$, tr$(A^{t}\overline{I_{n}})$=tr$(A^{t}I_{n})=0$, so $W^{⊥}=\{\lambda I_{n}|\lambda \in \mathbb{C}\}$, where $I_{n}$ is the identity matrix of dimension $n$.
A: Note that
$$
\operatorname{tr}(A^T \bar B) = 
\sum_{j=1}^n \sum_{k=1}^n a_{jk} \overline{b_{jk}} = 
\sum_{j=1}^n a_{jj}\overline{b_{jj}} + \sum_{j \neq k} a_{jk}\overline{b_{jk}}
$$
Now, fix any $B$ such that that $\langle A,B \rangle = 0$ for every $A \in W$.
Set $A_1 = (a_{jk})$ where
$$
a_{jk} = 
\begin{cases}
b_{jk} & j \neq k\\
0 & j = k
\end{cases}
$$
Because $A_1 \in W$, we have $\langle A_1,B \rangle = 0$.  What does this tell you about the off-diagonal entries of $B$?
Now, for any $p \neq q$ from $1$ to $n$, set $A_{pq} = (a_{jk})$ where
$$
a_{jk} = 
\begin{cases}
1 & j = k = p\\
-1 & j = k = q\\
0 & \text{otherwise}
\end{cases}
$$
What does this tell you about the diagonal entries of $B$?
Conclude that $W^\perp = \{\lambda I: \lambda \in \mathbb{C}\}$, where $I$ is the identity matrix.
