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$V$ is the matrices space (scalar over the complex).

definition of inner product space is:

$(A,B)=tr(AB^*)$. $A$,$B$ matrices.

assuming $D$ is the subspace of all Diagonal matrices.

I need to find the subspace that each matrix $B$ in it

$(A,B)=0$. $A$ belongs to $D$.

$[A]_{i,j}=a_{i,j},\quad [B]_{i,j}=b_{i,j}$

so I figure it out that the $tr(AB^*)= \sigma(a_{i,i}c_{i,i}),\quad 1\leq i\leq n$

while $c$ is the conjugation of $b$.

but compare it to zero still do not give any information what $c_{i,i}$ is.

Cameron Buie
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wantToLearn
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    I think you are trying to find the "orthogonal complement" of the space of diagonal matrices. There are standard techniques for that in the textbooks and on the web. – Gerry Myerson May 15 '13 at 13:57
  • Hint: If you pick a basis B1, B2, ..., Bk of D, then A will be in the orthogonal complement of D if and only if tr(A Bi*) = 0 for all i. Is there a nice basis of D to use? – Alex Zorn Mar 18 '15 at 15:42

3 Answers3

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Note that "orthogonal complement" means a subspace in which every vector is orthogonal to every vector in your given subspace. So in order that $D$ is a orthogonal complement to diagonal matrices, for any matrix $B \in D$ you'll get $$ \sum_i A_{ii} \overline{B}_{ii} = 0 $$ which has exactly the same form standard inner product over $C^n$, for diagonal vectors of $A$ and $B$.

Now since this holds for arbitrary $A$ with diagonal vector $(A_{11}, \cdots, A_{nn})$, the diagonal vector for $B$ has to be the zero vector, based on the result we obtain from $C^n$. Therefore, the subspace could be described as all matrices with diagonal entries $0$.

Jiageng
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By vectorising matrices in $M_n(\mathbb C)$ into vectors in $\mathbb C^{n^2}$, the inner product $\langle A,B\rangle=\operatorname{tr}(AB^\ast)$ is just the usual inner product $\langle \operatorname{vec}(A), \operatorname{vec}(B) \rangle = \operatorname{vec}(B)^\ast \operatorname{vec}(A)$ for vectors. So, the orthogonal complement in question is essentially $\operatorname{vec}(D)^\perp$. Repacking vectors in this subspace into matrices, you get the answer.

user1551
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By $E_i$ denote the $n\times n$ matrix whose elements are zero except that the element at the $i$th row and $i$th column is 1. Note $E_i$'s are orthogonal, because $tr(E_iE_j^*)=tr(\mathbf{0})=0$ for $i\neq j$. So $\{E_1, ..., E_n\}$ are linearly independent and $D=span(\{E_1, ..., E_n\})$. Therefore $D^\bot$ is the space of all matrices $A$ that $tr(AE_i^*)=0$ for all $i$. Note $tr(AE_i)=A_{ii}$. So $A\in D^\bot$ iff $A_{ii}=0$ for all $i$.

Mohsenhd
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