Subrepresentations of the tensor square I've been reading Serre's Book on Representation Theory, and in chapter 1.6 he introduces us to the symmetric space and alternating space. In the algebra course I did we covered tensor products, and we didn't talk about these two subspaces, so i'm having some doubts about them.
He starts by introducing an automorphism of $V \otimes  V$ : 
$$\theta(x\otimes y) = y \otimes x .$$ Then the space decomposes into $$\operatorname{Sym}^2(V) \oplus \operatorname{Alt}^2(V)$$ somehow; I don't really see where he gets this from the automorphism. Then he claims that $\operatorname{Sym}^2(V)$ has dimension $n(n+1)/2$ and that $\operatorname{Alt}^2(V)$ has dimension $n(n-1)/2$ when $V$ has dimension $n$. I don't really get where this comes from either so any help is appreciated.
 A: Think in terms of spaces of matrices. The space of $n\times n$ matrices is of dimension $n^2$ and represents $V\otimes V$. The automorphism introduced is equal to the matrix transpose. The subspace fixed by the automorphism is the space of symmetric matrices. Since a symmetric matrix is uniquely determined by its values on and above the diagonal, it has dimension $\binom{n+1}{2}$ (since that is the number of such entries). Any matrix can be uniquely decomposed as the sum of a symmetric matrix and an antisymmetric matrix, and the antisymmetric matrices have dimension $\binom{n}{2}$ since they are uniquely determined by the entries above the diagonal (the diagonal being equal to $0$).
A: Hint Since $\theta^2 = 1$, $\theta$ is diagonalizable, and its possible eigenvalues are $\pm 1$. By construction (or by definition, if we like) $\operatorname{Sym}^2 V$ and $\operatorname{Alt}^2 V$ are respectively the $(+1)$- and $(-1)$-eigenspaces of $\theta$, and so $$V \otimes V = \operatorname{Sym}^2 V \oplus \operatorname{Alt}^2 V$$ as claimed.

Additional hint The projections onto each eigenspace are \begin{aligned}\operatorname{Sym} : &V \otimes V \to \operatorname{Sym}^2 V, & \quad & A \mapsto \frac{1}{2}(A + \theta(A)) \\ \operatorname{Alt} : &V \otimes V \to \operatorname{Alt}^2 V, &\quad & A \mapsto \frac{1}{2}(A - \theta(A)) .\end{aligned} To compute the dimensions of the eigenspaces, fix a basis $(E_a)$ of $V$ and compute the image of the induced basis $(E_a \otimes E_b)$ of $V \otimes V$ under each of the maps. Then, for example, $\operatorname{Sym}^2 V = \operatorname{Sym}(V \otimes V)$ is spanned by $\operatorname{Sym}(E_a \otimes E_b) = \frac{1}{2} (E_a \otimes E_b + E_b \otimes E_a)$, and so the set $$\{\frac{1}{2} (E_a \otimes E_b + E_b \otimes E_a) : a \leq b\}$$ spans $\operatorname{Sym}^2 V$. It is linearly independent, so if $V$ has finite dimension, say, $n$, the dimension of $\operatorname{Sym}^2 V$ is the number of pairs $(a, b)$ of integers satisfying $1 \leq a \leq b \leq n$, and there are $\frac{1}{2} n (n - 1)$ of these. Of course, once you know the dimension of on of $\operatorname{Sym}^2 V$ and $\operatorname{Alt}^2 V$ you know the other, since $$\dim \operatorname{Sym}^2 V + \operatorname{Alt}^2 V = \dim (\operatorname{Sym}^2 V \oplus \operatorname{Alt}^2 V) = \dim(V \otimes V) = (\dim V)^2 = n^2 .$$

