Orthogonal matrices and inner product definition Let $n \geq 2$ and let $W$ be the subspace of 
$\mathbb M_n(\mathbb R)$ consisting of all matrices whose trace is zero. If $ A = (a_{ij})$ and $B = (b_{ij})$, for  $1 \leq$ i,j $\leq n$ , are elements in $\mathbb M_n(\mathbb R)$, define their inner-product by $$(A,B) = \sum_{i,j=0}^n a_{ij}b_{ij}$$ Identify the subspace $W^\perp$ of elements orthogonal to the subspace W.
My work : I am actually confuse in the symbols and on the basis of what I understand, I took a $2\times 2$ then $3\times 3$ matrices then i reached to the conclusion that "if i take two $n\times n$ matrices A and B with entries $a_{11},a_{12}\ldots ,a_{1n},\ldots, a_{nn}$  $b_{11},b_{12},\ldots, b_{1n},\ldots, b_{nn}$
Then inner product is $a_{11}b_{11}+a_{12}b_{12}+....a_{nn}b_{nn}$ " (here please check if I am thinking def right ?)
But after I equate it to zero and use trace for one matrix A (since other will be in $W^⊥$ I it will not have trace necessary zero so I took a random matrix B)
Then $$a_{11}+a_{22}+ \ldots +a_{nn}=0$$
Then I can have value of any one of $a_{ij}$ and put it there in that def of inner product but what do I have to do after that? All $b_{nn}$ are not necessarily non-zero  and all $a_{nn}$ are also not necessarily non-zero so I am not sure if I can take them comman and compare both sides and will get something useful and in this step how the def of  $W^\perp$ exactly looks like?
I'm not sure the way I'm thinking is right or not?
 A: Recall that the trace map is the function $\DeclareMathOperator{tr}{tr}\tr:\Bbb M_n(\Bbb R)\to\Bbb R$ given by $\tr(A)=\sum_k a_{kk}$. One easily shows that $\tr$ is linear. In this problem, we are interested in the kernel of $\tr$, which you call $W$ but I will refer to as $\mathfrak{sl}_n(\Bbb R)=\ker(\tr)$. Since $\tr$ has rank one, the rank-nullity theorem implies that 
$$
\dim\mathfrak{sl_n}(\Bbb R)=\dim\Bbb M_n(\Bbb R)-\dim\DeclareMathOperator{image}{image}\image(\tr)=n^2-1
$$
Now, we wish to describe $\mathfrak{sl}_n(\Bbb R)^\perp$ where we are viewing $\Bbb M_n(\Bbb R)$ as an inner-product space with inner product given by
$$
\langle A, B\rangle=\sum_{i,j}a_{ij}b_{ij}
$$
The dimension formula for orthogonal complements in finite-dimensional vector spaces implies that
$$
\dim\mathfrak{sl}_n(\Bbb R)^\perp=\dim\Bbb M_n(\Bbb R)-\dim\mathfrak{sl}_n(\Bbb R)=n^2-(n^2-1)=1
$$
Thus $\mathfrak{sl}_n(\Bbb R)^\perp$ is a one-dimensional subspace of $\Bbb M_n(\Bbb R)$. Can you compute a basis of $\mathfrak{sl}_n(\Bbb R)^\perp$?
A: We have one choice that is we can use it as : since $a_{11}+a_{22}+\ldots+a{nn}=0$ then $k(a_{11}+a_{22}+\ldots+a{nn}=0)$ so any matrix B which take only diagonal entries of a matrix A and just scale it and don't multiply with other entries of A then B should be a scalar matrix.
Other thing is if I put values of tr(A)=0 in that equation which I mention in question of inner product (and suppose all $a_{ij} \not = 0$ ) then I got $b_{ii}=b_{jj}$ for all i,j, and other entries zero so it's a scalar matrix . 
"Now here I took all $a_{ij} \not = 0$ in this point i'm not sure."
Note : now I'm sure of the definition so don't worry about that. :)
