# what is dimension of orthogonal complement of a subspace of a vector space.

This is last part of my other question. I don't understand the last part of problem. Feel free to edit the question.

c) Let $V$ be a vector space of real $n \times n$ symmetric matrices, what is $\text{dim } V$? What is the dimension of the subspace of $W$ of $V$ containing matrices $A$ whose trace is $0$? What is the dimension of the orthogonal complement of $W^{\perp}$ relative to the positive definite scalar product of part $(b)$?

Here is screen shot of the complete question

I think the $\text{dim } V = \frac{n(n+1)}{2}$ and $\text{dim} W = \frac{n(n+1)}{2} - 1$. Since one element of diagonal can be expressed linear combination of other diagonal elements. I don't understand the last part of it. dimension of the orthogonal complement of $W^{\perp}$.

"Complement", with an "e", and "compliment", with an "i", are two different things. If I say this question is brilliant, that's a compliment (with an "i"). A complement (with an "e") of $X$ is something that when added to $X$, makes a complete whole. The similarity between the spelling of "complement" (with an "e") and the spelling of "complete" is not coincidental and can serve as a reminder. I corrected the spelling in the question. –  Michael Hardy Jan 31 '13 at 15:11
In your case, $W^{\perp} = \{ X \in V : \text{$\operatorname{tr}(A X) = 0$for all$A \in W$} \}$. –  Andreas Caranti Jan 31 '13 at 15:16
@AndreasCaranti is $X$ vector or another matrix? –  hasExams Jan 31 '13 at 15:42
Since it is a complement, we have $\dim(W) + \dim(W^{\perp}) = \dim(V)$ –  David Wheeler Jan 31 '13 at 16:39