Find eigenvectors of antisymmetric matrix

Question

How I can found the eigenvectors in the matrix A?

$$ A= \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \\ \end{bmatrix} $$ with the condition $a^2+b^2+c^2=1$

I have the Characteristic polynomial: $ \lambda \ ( - \lambda \ ^2-1)$ and the eigenvalues $ \lambda \ _1 =0$, $ \lambda \ _2 =i$, $ \lambda \ _3 =-i$

My idea is use the definition for example for $ \lambda \ _2 =i$

$(A- \lambda \ _2 I ) (x,y,z)^T =(0,0,0)^T$

then i get the system $$\begin{equation}\begin{aligned} -ix+cy-bz = 0 \\ -cx-iy+az= 0 \\ bx-ay-iz= 0\\ \end{aligned}\end{equation}\tag{1}\label{eq1}$$ I don't see a solution to find the eigenvectors, can you gime some hints?

Answer 1

For a $3\times 3$ matrix $A$ with eigenvalues $\{\lambda_1,\lambda_2,\lambda_3\}\,$ Cayley-Hamilton tells us that $$p(A) = (A-\lambda_1I)(A-\lambda_2I)(A-\lambda_3I) = 0$$ Define the vectors $$\eqalign{ x_1 &= (A-\lambda_2I)(A-\lambda_3I)b_1 \cr x_2 &= (A-\lambda_3I)(A-\lambda_1I)b_2 \cr x_3 &= (A-\lambda_1I)(A-\lambda_2I)b_3 \cr }$$ where $\,b_k\,$ is any vector which makes $\,x_k\,$ non-zero.

By direct calculation $$\eqalign{ (A-\lambda_1I)x_1 &= p(A)b_1 = 0 \cr (A-\lambda_2I)x_2 &= p(A)b_2 = 0 \cr (A-\lambda_3I)x_3 &= p(A)b_3 = 0 \cr }$$ Thus $x_k$ is the eigenvector associated with eigenvalue $\lambda_k$

Answer 2

Given the Levi-Civita tensor $\varepsilon$ and the unit vector $n = [\,a\,\,b\,\,c\,]^T$, you can construct the matrix $$A=-\varepsilon\cdot n$$ This matrix has many interesting properties, among them $$\eqalign{ A^T &= -A \cr Ab &= n\times b \cr An &= n\times n = 0 \cr }$$ That last equation reveals that $n$ is the eigenvector associated with the eigenvalue $\lambda_1=0$.

The first few powers of $A$ are $$\eqalign{ A^2 &= nn^T-I \cr A^3 &= -A \cr }$$ Let $C_2=(A^2+\lambda_2A)\,$ then for an arbitrary vector $b$ we can calculate $$\eqalign{ x_2 &= C_2b = A(Ab+ib) \cr Ax_2 &= A^3b+iA^2b \cr &= i^2Ab+iA^2b \cr &= iA(ib+Ab) \cr &= ix_2\cr }$$ Therefore $x_2$ is the eigenvector associated with the eigenvector $\lambda_2=i$.

A similar calculation for $C_3=(A^2+\lambda_3A)\,$ shows that $\,x_3=C_3b$
is the eigenvector associated with $\lambda_3=-i$.