Prove that the eigenvalues of skew-Hermitian matrices are purely imaginary I would like some help on proving that the eigenvalues of skew-Hermitian matrices are all pure imaginary. I have gotten started on it, but am getting stuck.
Attempt at proof: $Av=\lambda v \implies A \bar{v}=\bar{\lambda}\bar{v}.$
Also, $v^TA^T=\lambda v^T \implies v^TA^T\bar{v}=\lambda v^T \bar{v} \implies ?$
Should I conjugate both sides next?
 A: We first of all notice that $A$ is normal, as $A^\dagger=-A$ implies $[A,A^\dagger]=0$, and therefore unitarily diagonalisable.
Write its eigendecomposition as $A=V\Lambda V^\dagger$ for some unitary $V$ and diagonal $\Lambda$. Then, $A^\dagger=V\Lambda^* V^\dagger$, and therefore $A^\dagger=-A\Longleftrightarrow\Lambda^*=-\Lambda$, which implies that the eigenvalues are purely imaginary.
A: Let $x$ an eigenvector of $A$ associated to the eigenvalue $\lambda$ then 
$$\langle Ax,x\rangle=\overline \lambda||x||^2=\langle x,A^*x\rangle=-\langle x,Ax\rangle=-\lambda\langle x,x\rangle$$
so 
$$\overline\lambda=-\lambda$$
hence $\lambda$ is pure imaginary complex.
A: Consider the Euclidean inner product of $x,y$ $\in$ Complex vectors $C^n$    defined as : $$\langle x,y \rangle = x^Hy$$
Where $x^H$ is the hermitian transpose of x.We also know $\overline{\lambda}$ is eigenvalue of $x^H$: $$Ax = \lambda x$$
$$(Ax)^H = (\lambda x)^H$$
$$x^H A^H = \overline{\lambda} x^H$$
Now:
$$\langle Ax,x \rangle = (\lambda x)^H x = \overline{\lambda} x^H x$$
Also :
$$\langle Ax,x \rangle = (Ax)^H x = x^H A^H x = x^H(A^Hx)$$
Since we know $A^H = -A$:
$$x^H(A^Hx) = x^H(-Ax) = - x^H(\lambda x) = -\lambda x^H x$$
Therefore:
$$-\lambda =  \overline{\lambda}$$
And $\lambda$ is purely imaginray.
A: $AX=λX$
so $X^H AX=X^H λX$
$-A^H X=λX$, so $X^H A^H X=-X^H λX$
$X^H AX=-X^H λ^H X$
Hence, $λ=-λ^H$
