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What will be the eigenvalues of the antisymmetric matrix $ A=\begin{pmatrix} 0 & -n_3 & n_2 \\n_3 & 0 & -n_1 \\-n_2 & n_1 & 0\end{pmatrix} $, where $ n_1$, $n_2$ & $n_3 $ are the components of a unit vector?

I need the way to solve it.

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What did you try? What about the standard way $A-\lambda1=0$? –  draks ... Oct 8 '12 at 10:29
    
oh I missed the determinant: $\det(A-\lambda1)=0$ –  draks ... Oct 8 '12 at 10:36

1 Answer 1

up vote 2 down vote accepted

Hint: Your matrix being a antisymmetric of odd order, should have $0$ as an eigenvlaue (why?). Now from the trace condition, you see that the remaining two have opposite sign. So, you need to calculate only the coefficient of $\lambda$ in the characteristic equation, which is sum of the three $2\times2$ principle minor. If you calculate it and use your condition $|\vec{n}|^2=1$, it will be a very well known number....

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