how to find the dimension of $ W^{\perp}$?

Question

For $ M_n( \mathbb{C})$ ,the vector space of $all \ n \times n$ complex matrices over $\mathbb{C}$

Given that $W =\{ x \in M_n( \mathbb{C}| trace(X) =0\}$ is a subspace of $ M_n( \mathbb{C})$ and $W ^{\perp}$ consists of sacalar matrices that is if $tr(AX) = 0$ for all $x \in M_n( \mathbb{C})$ with $trX =0$ ,then $A= \lambda I$ for some scalars $\lambda$ . Find the dimension of

$1)$$ W$

$2)$$W^{\perp}$

My attempts :

i know that dimension of $M_n(\mathbb{C})$ is $2n^2.$ now $trace A =a_{11} +a_{22} +a_{33} +......+a_{nn}=0$

so the dimension of $W = 2n^2 -1$

here im confusion about that how to find the dimension of $ W^{\perp}$?

any hints/solution will be appreciated

thanks u

Answer 1

The dimension of $M_n(\mathbb{C})$ is $n^2$, not $2n^2$. And anyway, if you know how to find the dimension of $W$ then the dimension of $W^{\perp}$ is the codimension of $W$, which means $dim(M_n(\mathbb{C}))-dim(W)$.