Find eigenvectors of antisymmetric matrix 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?
 A: 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 $\lambda_3=-i$.
A: 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 $\lambda_k$
