Question: Let $V \subset M(n,n,\mathbb{R})$ be the set of all symmetric, real $(n \times n)$ matrices, that is $a_{ij} = a_{ji}$ for all $i,j$. Show that $V$ is a subspace of $M(n,n,\mathbb{R})$ and calculate dim$(V)$.
My attempt so far: First part: To show that $V$ is a subspace I need to show: (a) $ 0 \in V$ and (b) $\forall A,B \in V: (i) A + B \in V (ii) \lambda A \in V$
For (a) I would say: Let $a_{ij} \in 0$(this should represent a zero matrix, is that how to write it?)
$a_{ij} = 0 = a_{ji} \Rightarrow 0 \in V$
For (b) I am actually confused since I would first think: both a $(2 \times 2)$ matrix and a $(3 \times 3)$ matrix belong to $V$ but addition of matrices of different size is undefined $\Rightarrow$ $V$ is not closed under addition $\Rightarrow$ $V$ is not a subspace of $M(n,n,\mathbb{R})$... what am I missing here? (To start I don't really understand the notation $M(n,n,\mathbb{R})$... what exactly does the $\mathbb{R}$ represent there?).
Disregarding my confusion I would still try to show (b), but my mathematical notation is still lacking competence... Is the following somewhat clear? Would anyone ever use "$\in$" to denote "is an element of matrix"?
(i)Let $a_{ij},a_{ji} \in A$ and $b_{ij}, b_{ji} \in B$. Let $A,B \in V$
$\Rightarrow a_{ij} = a_{ji}, b_{ij} = b_{ji}$
$A + B = C \Rightarrow c_{ij} = (a_{ij}+b_{ij}) = (a_{ji} + b_{ij}) = (a_{ij} + b_{ji}) = c_{ji} = (a_{ji} + b_{ji})$
$\Rightarrow C \in V$
(ii) Let $A\in V, \lambda \in \mathbb{R}$. Let $a_{ij},a_{ji} \in A$.
$\Rightarrow a_{ij} = a_{ji}$
$\lambda \cdot A = A'$ with $\lambda a_{ij} = \lambda a_{ji} \Rightarrow A' \in V$
Second part: I feel that I understand the answer... For an $(n \times n)$ matrix, the diagonal length $ = n$ and these are the elements which have no counterpart and are not critical to the symmetry. When these elements are subtracted from the total$(n^{2})$, half of the remainder can be independently selected and the other half will follow as a result. Therefore I think it makes sense to write that dim$(V) = n + \frac{n^{2}-n}{2}$.
Is this correct? If so, given the context of the exercise, how could I make my answer more acceptable?