These types of matrices are invertible I was reading a paper and the author asserts this (for finite dimensional vector spaces):

It is not difficult to prove the sum of a positive definite operator plus a
  skew-symmetric one is inversible.

Can you help me to see why this is true?
 A: If $V$ is a finite-dimensional inner product space, then I shall use the following terminology:


*

*An endomorphism $A$ of $V$ is said to be positive definite if every $v \in V$ satisfies $\left<v, Av\right> \geq 0$, with equality holding only for $v = 0$.

*An endomorphism $A$ of $V$ is said to be skew-symmetric if every $v \in V$ and $w \in V$ satisfy $\left<Av, w\right> + \left<v, Aw\right> = 0$.
(When $V$ is the $\mathbb{R}$-vector space $\mathbb{R}^n$ equipped with its usual inner product (the dot product), these definitions of "positive definite" and "skew-symmetric" are equivalent to the usual definitions of "positive definite" and "skew-symmetric" for matrices, if you identify endomorphisms of $\mathbb{R}^n$ with $n\times n$-matrices, as long as your definition of a "positive definite matrix" does not require the matrix to be symmetric. However, if $V$ is $\mathbb{R}^n$ equipped with a different inner product, these definitions can be something rather different.)
Now, I am interpreting your question as follows:

Let $V$ be a finite-dimensional inner product space. Let $P$ be a positive definite endomorphism of $V$, and let $S$ be a skew-symmetric endomorphism of $V$. Prove that $P + S$ is invertible.

Proof. Let $v \in \operatorname{Ker}\left(P + S\right)$. Thus, $v \in V$ and $\left(P + S\right) v = 0$. Since $S$ is skew-symmetric, we have $\left<Sv, v\right> + \left<v, Sv\right> = 0$, thus $-\left<v, Sv\right> = \left<Sv, v\right> = \left<v, Sv\right>$ (since the inner product is symmetric). Hence, $2 \left<v, Sv\right> = 0$, and thus $\left<v, Sv\right> = 0$. But
$0 = \left<v, \underbrace{0}_{= \left(P + S\right) v}\right> = \left< v, \left(P + S\right) v \right>$
$= \left< v, Pv + Sv \right> = \left< v, Pv\right> + \underbrace{\left< v, Sv\right>}_{=0} = \left< v, Pv\right>$.
But $P$ is positive definite. Hence, $\left< v, Pv\right> = 0$ can only happen when $v = 0$. Thus, we have $v = 0$ (since $\left< v, Pv\right> = 0$).
Now, forget that we fixed $v$. We have thus shown that $v = 0$ for every $v \in \operatorname{Ker}\left(P + S\right)$. In other words, $\operatorname{Ker}\left(P + S\right) = 0$. Hence, the linear map $P + S$ is injective. Since $P + S$ is an endomorphism of the finite-dimensional vector space $V$, this entails that $P + S$ is invertible. This completes the proof.
Remark: Note that we have only used the $\mathbb{Z}$-bilinearity and the symmetry of the form $\left<\cdot, \cdot\right>$, not its positive definiteness. Thus, our inner product space could be replaced by any $\mathbb{R}$-vector space equipped with a $\mathbb{Z}$-bilinear form.
