Find the Jordan canonical form of a $3\times 3$ antisymmetric matrix

Question

$\left[ {\begin{array}{c} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{array} } \right]$ where $a,b,c \in \mathbb{R}$

What is the Jordan canonical form of this matrix? I'm not sure how to do this. Any solutions/help is greatly appreciated

Answer 1

Compute the characteristic polynomial, and its roots, those are your eigenvalues. In that case, you will find $0$ and two imaginary roots that are conjugate to each other, so your matrix is diagonalizable.

Answer 2

Its characteristic polynomial is: $$\chi_A(\lambda)=-\bigl(\lambda^3+(a^2+b^2+c^2)\lambda\bigr)=-\lambda(\lambda^2+a^2+b^2+c^2)$$ and it does not split into linear factors, hence $A$ is not triangularisable over $\mathbf R$.

Over $\mathbf C$ it splits into three distinct linear factors $$\chi_A(\lambda)=-\lambda\bigl(\lambda+\mathrm i\sqrt{a^2+b^2+c^2}\bigr))\bigl(\lambda-\mathrm i\sqrt{a^2+b^2+c^2}\bigr),$$ hence $A$ is even diagonalisable over $\mathbf C$ as $$\begin{bmatrix} 0&0&0\\0&-\mathrm i\sqrt{a^2+b^2+c^2}&0\\ 0&0&\mathrm i\sqrt{a^2+b^2+c^2} \end{bmatrix}$$

Answer 3

The matrix is diagonalizable with eigenvalues $\lambda_1=0$ and $\lambda_{1,2}=\pm\sqrt{-(a^2+b^2+c^2)}$ that are solutions of the equation: $$ \lambda^3+\lambda(a^2+b^2+c^2)=0 $$ so the Jordan canonical form is the diagonal matrix: $$ \begin{bmatrix} 0&0&0\\ 0&\sqrt{-(a^2+b^2+c^2)}&0\\ 0&0&-\sqrt{-(a^2+b^2+c^2)} \end{bmatrix} $$

(find the eigenvectors is a bit more work).