Find the Jordan canonical form of a $3\times 3$ antisymmetric matrix $\left[ {\begin{array}{c}
 0 & a & -b \\
 -a & 0 & c \\
 b & -c & 0
 \end{array} } \right]$
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
 A: 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
$$\begin{bmatrix}
0&0&0\\0&-\mathrm i\sqrt{a^2+b^2+c^2}&0\\
0&0&\mathrm i\sqrt{a^2+b^2+c^2}
\end{bmatrix}$$
A: 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.
A: 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).
