# Eigenvalues of a real skew symmetric matrix

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 $$\iff\langle x,Ax\rangle =0\,,\forall x$$.