\mbox{How to find the eigenvalue of 2x2 skew-symmetric matrix} \\ \begin{aligned} \mathbf{M}= \begin{bmatrix} 0 & 4\\ -4 & 0 \end{bmatrix} \end{aligned}\\ \mbox{Steps to find the eigenvalues of a matrix} \\ \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}
$\mbox{How to find the eigenvalues of 3x3 skew-symmetic matrix}$ \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} $\mbox{Apply cofactor expansion in the 3x3 matrix}$ \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}
$\mbox{Prove all non zero eigenvalues of real skew-symmetric matrices are pure imaginary}$ $\mathbf{A}^{\ast} = -\mathbf{A}$ $\mathbf{u}^{\ast} \mathbf{A}^{\ast} \mathbf{v} = (\mathbf{A} \mathbf{u})^{\ast} \mathbf{v}$ $\Rightarrow \langle \mathbf{A} \mathbf{u} \,, \mathbf{v} \rangle = \langle \mathbf{u} \,, \mathbf{A}^{\ast} \mathbf{v} \rangle$ $\Rightarrow \langle \mathbf{A} \mathbf{u} \,, \mathbf{v} \rangle = \langle \mathbf{u} \,, \mathbf{A}^{T} \mathbf{v} \rangle \quad A \mbox{ is real}$ $\mbox{Let } \mathbf{u} = \mathbf{v}$ $\Rightarrow \langle \mathbf{A} \mathbf{v} \,, \mathbf{v} \rangle = \langle \mathbf{v} \,, \mathbf{A}^{T} \mathbf{v} \rangle \quad A \mbox{ is real}$ $\mbox{Let } \mathbf{A}\mathbf{v} = \lambda \mathbf{v} \quad \mathbf{v} \neq 0 \quad \lambda \mbox{ is eigenvalue}$ $\Rightarrow \langle \lambda \mathbf{v} \,, \mathbf{v} \rangle = \langle \mathbf{v} \,, -\mathbf{A} \mathbf{v} \rangle$ $\Rightarrow \langle \lambda \mathbf{v} \,, \mathbf{v} \rangle = \langle \mathbf{v} \,, - \lambda\mathbf{v} \rangle$ $\Rightarrow \lambda \langle \mathbf{v} \,, \mathbf{v} \rangle = - \overline{\lambda}\langle \mathbf{v} \,, \mathbf{v} \rangle$ $\Rightarrow \lambda = - \overline{\lambda} \quad \because \mathbf{v} \neq \mathbf{0}$ $\Rightarrow \lambda \mbox{ is pure imaginary}$
\begin{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}