Given a real, skew-symmetric matrix $\mathbf{A}\in\mathbb{R}^{m\times m}$, and a nonzero vector $X\in\mathbb{R}^{m}$, classify the scalar $\lambda$ (real, complex, imaginary, etc.) in the eigenvalue equation $$ \mathbf{A}X = \lambda X $$
The matrix $\mathbf{A}$ is skew symmetric if $\mathbf{A}^{T}=-\mathbf{A}$.
My try : I am new to matrices , so can not get a idea how to deal with it , Any hint will help
Multiplying by $X^T$, you have $$ X^TAX=\lambda X^TX $$ Transposing gives $$ X^TA^TX=\lambda X^TX $$ Now use that $A^T=-A$.
$AX = \lambda X; \tag 1$
Now let us suppose we are dealing with real vectors $0 \ne X \in \Bbb R^n$, which is equipped with the usual inner product $\langle \cdot, \cdot \rangle$ such that
$\langle Y, AZ \rangle = \langle A^T Y, Z \rangle, \forall Y, Z \in \Bbb R^n, \tag 2$
which is essentially the definition of $A^T$ in terms of $\langle \cdot, \cdot \rangle$; then $\lambda \in \Bbb R$, since the entries of both $A$ and $X$ in (1) are real themselves, whence
$\bar \lambda X = \bar \lambda \bar X = \overline{\lambda X} = \overline{AX} = \bar A \bar X = AX = \lambda X, \tag 3$
or
$(\bar \lambda - \lambda)X = 0; \tag 4$
thus with
$X \ne 0, \tag 5$
$\bar \lambda - \lambda = 0 \Longleftrightarrow \lambda = \bar \lambda \Longleftrightarrow \lambda \in \Bbb R, \tag 6$
that is, $\lambda$ must be real. Furthermore,
$\lambda \langle X, X \rangle = \langle X, \lambda X \rangle = \langle X, AX \rangle = \langle A^T X, X \rangle$ $= \langle -AX, X \rangle = -\langle X, AX \rangle = -\langle X, \lambda X \rangle = -\lambda \langle X, X \rangle; \tag 7$
with $X \ne 0$, this forces
$\lambda = -\lambda \Longrightarrow \lambda = 0. \tag 8$
We may extend the vector space $\Bbb R^n$ on which $A$ acts to $\Bbb C^n$ in the usual manner, extending as well the inner product to $\langle \cdot, \cdot \rangle_{\Bbb C}$ in the usual sense so that
$\bar \mu \langle Y, Z \rangle_{\Bbb C} = \langle \mu Y, Z \rangle_{\Bbb C}, \; \mu \langle Y, Z \rangle_{\Bbb C} = \langle Y, \mu Z \rangle_{\Bbb C}, \; \mu \in \Bbb C,\; Y, Z \in \Bbb C^n; \tag 9$
if we leave $A$ unaltered, so that we still have
$\bar A = A, \tag{10}$
then the Hermitian adjoint of $A$ is
$A^\dagger = (\bar A)^T = A^T = -A, \tag{11}$
and $A$ may be regarded as a skew-Hermitian matrix, which as such satisfies the analog of (2):
$\langle Y, AZ \rangle_{\Bbb C} = \langle A^\dagger Y, Z \rangle_{\Bbb C}; \tag{12}$
then by virtue of (9), the analog of (7) reads
$\lambda \langle X, X \rangle_{\Bbb C} = \langle X, \lambda X \rangle_{\Bbb C} = \langle X, AX \rangle_{\Bbb C} = \langle A^\dagger X, X \rangle_{\Bbb C}$ $= \langle -AX, X \rangle_{\Bbb C} = -\overline{\langle X, AX \rangle}_{\Bbb C} = -\overline{\langle X, \lambda X \rangle}_{\Bbb C} = -\overline{\lambda \langle X, X \rangle}_{\Bbb C} = -\bar \lambda\overline{\langle X, X \rangle}_{\Bbb C} = -\bar \lambda \langle X, X \rangle_{\Bbb C}; \tag {13}$
again, $X \ne 0$ implies $\langle X, X \rangle_{\Bbb C} \ne 0$ and so
$\lambda = -\bar \lambda \Longleftrightarrow \bar \lambda = -\lambda \Longleftrightarrow \lambda \in i \Bbb R; \tag{14}$
that is, $\lambda$ is a pure imaginary number, which may vanish, but not necssarily.
Hint: $A$ is skew symmetric $\iff\langle x,Ax\rangle =0\,,\forall x$.
