Confusion in computing dimension of symmetric matrices when fields are changed Let $$V_1=\{A\in M_{n\times n}: a_{ij} \in \mathbb{C} \text{ and } A=A^\tau\}$$
Let $$V_2=\{A\in M_{n\times n}: a_{ij} \in \mathbb{R} \text{ and } A=A^\tau\}$$
I want to find out the dimension of $V_1$ over the field $\mathbb{R}$ and $\mathbb{C}$ and the dimension of $V_2$ over the field $\mathbb{R}$.

My confusion
I am clear that dimension of set on $n\times n$ symmetric matrices are $n(n+1)/2$ and how to determine its basis. But I am confused how the dimension would be changed if the field is changed. For exp dimension of $V_1$ as I know would be $n(n+1)$ when the field $F = \mathbb{R}$ and $n(n+1)/2$ when $F = \mathbb{C}$. But why is this changed of dimension?
For $V_2$ now entries $a_{ij} \in \mathbb{R}$ instead of from $\mathbb{C}$. Now I am confused how to compute its dimension? As what changes I have to see if the entries are now real instead of complex?
Edits
Notations
$^\tau$ :transpose
$\mathbb{R}$: Real number
$\mathbb{C}$: Complex number
Thanks it would be of great help to me
 A: The dimension of $V_1$ as a $\mathbb{C}$-vector space and $V_2$ as an $\mathbb{R}$-vector space are identical: $\frac{n(n+1)}{2}$.
On the other hand the dimension of $V_1$ as $\mathbb{R}$-vector space is two times the dimension of $V_1$ as $\mathbb{C}$-vector space.
Finally, it doesn't make sense to talk about $V_2$ as a $\mathbb{C}$-vector space, since it would not be a vector space (you would obtain complex numbers that are not in $\mathbb{R}$).
More generally, you multiply by two the dimension of the vector space if your $\mathbb{C}$-vector space becomes a $\mathbb{R}$-vector space (in the case of finite dimension). Because for each coordinates you had with your original $\mathbb{C}$-vector space, you need two dimensions to have the information using only real scalars: for that it is necessary to have a real part and an imaginary part for each coordinate where with a complex scalar the two are expressed with only one scalar, which doubles in reality the dimension.
Example : $\mathbb{C}$ can be viewed as a 1 dimensional $\mathbb{C}$-vector space which basis is simply the vector (1) (any non zero vector), however $\mathbb{C}$ as a $\mathbb{R}$-vector space is a 2 dimensional vector space which basis can be (1,i).
In your case, to have a basis for the $\mathbb{R}$-vector space, you can take the $\frac{n(n+1)}{2}$ matrix that compose the usual basis of the $\mathbb{C}$-vector space for symmetric matrices, and you need to add the $\frac{n(n+1)}{2}$ same matrices where you replaced the 1 by i inside.
