\[
\mbox{How to find the eigenvalue of 2x2 skew-symmetric matrix} \\
\begin{equation}
\begin{aligned}
\mathbf{M}= \begin{bmatrix}
0 & 4\\
-4 & 0
\end{bmatrix}
\end{aligned}
\end{equation}\\
\mbox{Steps to find the eigenvalues of a matrix} \\
\]
\begin{equation}
\begin{aligned}
& \det\,(\lambda \mathbf{I} - \mathbf{M}\,) = 0 \quad \lambda \in \mathbb{R} \\
& \det \bigg(
\begin{bmatrix}
\lambda & 0\\
0 & \lambda
\end{bmatrix} -
\begin{bmatrix}
0 & 4\\
-4 & 0
\end{bmatrix} \bigg)= 0 \\
& \det \bigg(
\begin{bmatrix}
\lambda & 4\\
-4 & \lambda
\end{bmatrix} \bigg) = 0 \\
& \Rightarrow \lambda^{2} + 16 = 0 \quad \mbox{Characteristic Polynomial}\\
& \Rightarrow \lambda = \pm 4i
\end{aligned}
\end{equation}
\[ \mbox{How to find the eigenvalues of 3x3 skew-symmetic matrix} \]
\begin{equation}
\begin{aligned}
& \det\,(\lambda \mathbf{I} - \mathbf{M}\,) = 0 \quad \lambda \in \mathbb{R} \\
& \det \bigg(
\begin{bmatrix}
\lambda & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 &\lambda
\end{bmatrix} -
\begin{bmatrix}
0 & a & b \\
-a & 0 & c \\
-b & -c & 5
\end{bmatrix} \bigg)= 0 \\
& \det \bigg(
\begin{bmatrix}
\color{red}{\lambda} & a & b \\
\color{red}{-a} & \lambda & c \\
\color{red}{-b} & -c & \lambda
\end{bmatrix} \bigg) = 0 \\
\end{aligned}
\end{equation}
\[ \mbox{Apply cofactor expansion in the 3x3 matrix} \]
\begin{equation}
\begin{aligned}
& \color{red}{\lambda}(-1)^{1+1}
\begin{bmatrix}
\lambda & c \\
-c & \lambda
\end{bmatrix} +
\color{red}{-a}(-1)^{1+2}
\begin{bmatrix}
a & b \\
-c & \lambda
\end{bmatrix} +
\color{red}{-b}(-1)^{1+3}
\begin{bmatrix}
a & b \\
\lambda & c
\end{bmatrix} = 0 \\
& \Rightarrow \lambda(\lambda^2 + c^2) + a(a\lambda + bc) - b(ac - \lambda b) = 0 \\
& \Rightarrow \lambda^3 + \lambda c^2 + \lambda a^2 + abc - abc + \lambda b^2 = 0 \\
& \Rightarrow \lambda^3 + \lambda(a^2 + b^2 + c^2) = 0 \\
& \Rightarrow \color{red}{\lambda = 0} \mbox{ is one of eigenvalues} \\
& \quad \mbox{If } \lambda \neq 0 \\
& \Rightarrow \lambda^2 + (a^2 + b^2 + c^2) = 0 \\
& \Rightarrow \color{red}{\lambda =\pm i\sqrt{a^2 + b^2 + c^2}} \\
& \mbox{Therefore, the eigenvalues of } 3 \times 3 \mbox{ skew-symmetic matrix } \\
& \lambda_1 = 0 \,, \quad \lambda_2 = -i\sqrt{a^2 + b^2 + c^2} \,, \quad \lambda_3 = i\sqrt{a^2 + b^2 + c^2} \\
\end{aligned}
\end{equation}
\[ \mbox{Prove all non zero eigenvalues of real skew-symmetric matrices are pure imaginary} \]
\[ \mathbf{A}^{\ast} = -\mathbf{A} \]
\[ \mathbf{u}^{\ast} \mathbf{A}^{\ast} \mathbf{v} = (\mathbf{A} \mathbf{u})^{\ast} \mathbf{v} \]
\[ \Rightarrow \langle \mathbf{A} \mathbf{u} \,, \mathbf{v} \rangle = \langle \mathbf{u} \,, \mathbf{A}^{\ast} \mathbf{v} \rangle \]
\[ \Rightarrow \langle \mathbf{A} \mathbf{u} \,, \mathbf{v} \rangle = \langle \mathbf{u} \,, \mathbf{A}^{T} \mathbf{v} \rangle \quad A \mbox{ is real}\]
\[ \mbox{Let } \mathbf{u} = \mathbf{v}\]
\[ \Rightarrow \langle \mathbf{A} \mathbf{v} \,, \mathbf{v} \rangle = \langle \mathbf{v} \,, \mathbf{A}^{T} \mathbf{v} \rangle \quad A \mbox{ is real}\]
\[ \mbox{Let } \mathbf{A}\mathbf{v} = \lambda \mathbf{v} \quad \mathbf{v} \neq 0 \quad \lambda \mbox{ is eigenvalue} \]
\[ \Rightarrow \langle \lambda \mathbf{v} \,, \mathbf{v} \rangle = \langle \mathbf{v} \,, -\mathbf{A} \mathbf{v} \rangle \]
\[ \Rightarrow \langle \lambda \mathbf{v} \,, \mathbf{v} \rangle = \langle \mathbf{v} \,, - \lambda\mathbf{v} \rangle \]
\[ \Rightarrow \lambda \langle \mathbf{v} \,, \mathbf{v} \rangle = - \overline{\lambda}\langle \mathbf{v} \,, \mathbf{v} \rangle \]
\[ \Rightarrow \lambda = - \overline{\lambda} \quad \because \mathbf{v} \neq \mathbf{0} \]
\[ \Rightarrow \lambda \mbox{ is pure imaginary}\]
\begin{equation}
\begin{aligned}
& \mbox{Prove all the eigenvalues of symmetric matrix are real} \\
& \mbox{Let } \mathbf{A}\mathbf{v} = \lambda \mathbf{v} \quad \mathbf{v} \neq 0 \quad \lambda \mbox{ is eigenvalue} \\
& \Rightarrow (\mathbf{A}\mathbf{v})^{\ast} = (\lambda \mathbf{v})^{\ast} \\
& \Rightarrow \mathbf{v}^{\ast}\mathbf{A}^{\ast} = \mathbf{v}^{\ast} \lambda^{\ast} \\
& \Rightarrow \mathbf{v}^{\ast}\mathbf{A} = \mathbf{v}^{\ast} \lambda^{\ast} \quad \because \mathbf{A} \mbox{ is symmetic} \\
& \Rightarrow \mathbf{v}^{\ast}\mathbf{A}\mathbf{v} = \mathbf{v}^{\ast} \lambda^{\ast} \mathbf{v} \\
& \Rightarrow \mathbf{v}^{\ast} \lambda \mathbf{v}= \mathbf{v}^{\ast} \mathbf{v} \quad \because \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \\
& \Rightarrow \langle \mathbf{v} \,, \lambda\mathbf{v} \rangle = \langle \mathbf{v} \,, \lambda^{\ast}\mathbf{v} \rangle \\
& \Rightarrow \overline{\lambda} \langle \mathbf{v} \,, \mathbf{v} \rangle = \lambda \langle \mathbf{v} \,, \mathbf{v} \rangle \quad \because
\langle \mathbf{x} \,, \lambda \mathbf{y} \rangle = \overline{\lambda} \langle \mathbf{x} \,, \mathbf{y} \rangle \\
& \Rightarrow \lambda = \overline{\lambda} \quad \because \mathbf{v} \neq \mathbf{0} \,, \langle \mathbf{v} \,, \mathbf{v} \rangle \neq 0 \\
& \Rightarrow \lambda \in \mathbb{R} \\
\end{aligned}
\end{equation}