$T=-T^{*}$, show that $T+\alpha I$ is invertible. Please don't answer the question. Just tell me if I am in the right direction. I should be able to solve this.
We are given $T=-T^{*}$, show that $T+\alpha I$ is invertibe for all real alphas that aren't zero.
What I did:
$det(T+\alpha I) = det(-T^{*}+\alpha I)=det(-\bar T+\alpha I) = \overline {det(-T+\alpha I)}$
And here I'm pretty much stuck. Am I in the right direction?
 A: Since $T=-T^*$, we have for any $x\in\mathbb{C}^n$ that $\mathrm{Re}(x^*Tx)=0$. Assume that $\alpha I+T$ (with real $\alpha\neq 0$) is singular, that is, there exists a nonzero $y\in\mathbb{C}^n$ such that $(\alpha I + T)y=0$ and hence $\alpha y=-Ty$. Multiplying with $y^*$ gives $\alpha y^*y=-y^*Ty$. We get a contradiction, since $\alpha y^*y$ is real and nonzero, while $\mathrm{Re}(y^*Ty)=0$.
A: copper.hat's comment was on point, in my book.
As long as $T^*$ is such that $\langle Tu, v \rangle = \langle u, T^*v \rangle$ (see Oria Gruber's remarks in the comments), we have that $\langle x, Tx \rangle$ is purely imaginary for all vectors $x$, since
$\langle x, Tx \rangle^* = \langle Tx, x \rangle = \langle x, T^*x \rangle = -\langle x, Tx \rangle, \tag{1}$
where the raised $*$ means complex conjugation when applied to numbers, as in $\langle x, Tx \rangle^*$ which occurs at the left of equation (1), and the operation ${} ^*$ which takes $T \to T^*$ when applied to operators such as $T$.  In any event, this means the eigenvalues of $T$ are purely imaginary, since if we have 
$Tv = \lambda v \tag{2}$
with $\Vert v \Vert = \langle v, v \rangle^{1/2}= 1$, then
$\lambda = \lambda \Vert v \Vert = \lambda \langle v, v \rangle = \langle v, \lambda v \rangle = \langle v, Tv \rangle. \tag{3}$
We now use the salutory fact that, for any scalar $\alpha$,
$Tv = \lambda v \Leftrightarrow (T + \alpha I)v = (\lambda + \alpha)v; \tag{4}$
since as we have seen the eigenvalues of $T$ are purely imaginary, by (4) those of $T + \alpha I$ for real $\alpha \ne 0$ all have non-vanishing real part; hence none can be zero; hence $T + \alpha I$ is nonsingular and and so invertible.  QED.
Hope this helps.  Best Wishes for the New Year,
and as always,
Fiat Lux!!!
