dimension of symplectic lie algebra The symplectic Lie algebra is defined as follows.
Let dim($V$)$=2n$. 
We define a skew symmetrical bilinear form $f$ on $V$ with the matrix $S:=$
$\left( \begin{array}{rrrr}
0 & I_n \\
-I_n & 0\\
\end{array}\right) $. 
$f(v,w):=v^TSw$ for $v,w \in V$. 
$\mathfrak{sp}$($V$)$:=$$\{g\in \hspace{0.15cm} $End$_\mathbb{K}(V):f(g(v),w)=-f(v,g(w))$ for $v,w\in V$} 
 is a subalgebra of $\mathfrak{gl}$($V$), the sympletctic Lie Algebra
My question is why do we set the dimension of $V$ to be $2n$ and not just $n$?
Is it just a matter of definition or is there a true reason behind it?
 A: To expand the comments by user WE Tutorial School into an answer:
First of all, regarding your comment, for the orthogonal Lie algebra (which happens to be also the special orthogonal Lie algebra), it is not true that it exists only in odd dimension. Rather, it is just so that in even dimension $2n$, one uses the form given by the matrix $\pmatrix{0&I_n\\I_n&0}$, whereas in odd dimension $2n+1$, one uses the one given by $\pmatrix{1&0&0\\0&0&I_n\\0&I_n&0}$.
To be more precise, one could also replace these matrices $S$ by any ones that are congruent to them, because that gives the same symmetric bilinear form, just expressed in a different basis. In particular, over an algebraically closed field, where all symmetric forms are the same via base change, you could also just work with the identity matrix $S=I_m$ in any dimension $m$. Over other fields (like $\mathbb R$) though, one has as many non-isomorphic Lie algebras here as there are non-equivalent symmetric bilinear forms, as there are non-congruent symmetric matrices. For example over $\mathbb R$, if one takes for $S$ the identity matrix, one gets the "compact" form of the Lie algebras instead, which differ from the ones above (for $m\ge 3$ at least). The above $S$ with the slightly different definition in odd and even dimensions instead give the split forms of the (special) orthogonal Lie algebras.
The reason one often chooses those matrices, even over an algebraically closed field where at first $S=I_m$ looks like a more generic choice, is that if one writes out the matrices that make up the Lie algebra with respect to those parity-depending matrices $S$, it's relatively easy to "see" a nice Cartan subalgebra, root spaces etc. It matches that the root systems eventually look different for the odd and even dimensional case: For even $m=2n$, one gets a root system of type $D_n$, whereas for odd $m=2n+1$, one gets a root system of type $B_n$. For example here I recently worked with that for the case $n=2, m=2n+1=5$.
Now, to your actual question: as for the symplectic Lie algebra, why does it not work in odd dimension? Well it's a plain fact of linear algebra that on such spaces, any skew-symmetric bilinear form is degenerate. Note that the $S$ written down above for the orthogonal case had to be symmetric, $S^t=S$, and both in the odd and even dimension case the matrix given there satisfies that. But here we would need one with $S^t=-S$. Well in even dimension $m=2n$, $S=\pmatrix{0&I_n\\-I_n&0}$ does the job (and again gives a good matrix representation of the elements, eventually leading to the root system $C_n$; again, over non-algebraically closed field, there are in general other forms as well, and this one is just the "split form"). But in odd dimension, we cannot imitate the above trick: The attempt $\pmatrix{1&0&0\\0&0&I_n\\0&-I_n&0}$ (or any $\pmatrix{a\neq 0&0&0\\0&0&I_n\\0&-I_n&0}$) would not be skew-symmetric, whereas $\pmatrix{0&0&0\\0&0&I_n\\0&-I_n&0}$ of course is degenerate. And as said above, one shows in linear algebra that there is no choice of matrix in odd dimensions which gives what one would want. See the link given by WE Tutorial School, https://math.stackexchange.com/a/3629615/96384, or any good introduction to bilinear forms.
