4
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The possible set of eigenvalues of a $4\times 4$ Real skew symmetric, orthogonal matrix is

$1.\{\pm i\}$

$2.\{\pm i,\pm 1\}$

$3.\{\pm 1\}$

$4.\{\pm i,0\}$

As it is real skew symmetric so eigenvalues may be $0$ or Purely Imaginary, and as it is orthogonal so determinant must be $1$ or $-1$. So $1$ may be the possible set. Am I right?

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3
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Skew symmetric real matrices (more generally skew-Hermitian complex matrices) have purely imaginary (complex) eigenvalues. Orthogonal real matrices (more generally unitary matrices) have eigenvalues of absolute value$~1$. Either type of matrix is always diagonalisable over$~\Bbb C$. This means that for the given matrix only $\mathbf i$ and $-\mathbf i$ are possible eigenvalues. Moreover for real matrices the multiset of roots of the characteristic polynomial is invariant under complex conjugation; all this implies that $\mathbf i$ and $-\mathbf i$ are both eigenvalues, each with (geometric and algebraic) multiplicity$~2$.

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0
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Since $det(A−\lambda I)=det(A'−\lambda I)$, A and $A'$ have the same eigenvalues. On the other hand, $A'$ and −A also have the same eigenvalues. Thus if $\lambda$ is an eigenvalue of A, so is $-\lambda$. If n is odd, $\lambda=0$ is an eigenvalue.

If $\lambda$ is an eigenvalue of a matrix A, then $1/\lambda$ is an eigenvalue of $A^{-1}$.

Hence, ${\pm i}$ is the answer.

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-1
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0 down vote

Since det(A−λI)=det(A′−λI)det(A−λI)=det(A′−λI), A and A′A′ have the same eigenvalues. On the other hand, A′A′ and −A also have the same eigenvalues. Thus if λλ is an eigenvalue of A, so is −λ−λ. If n is odd, λ=0λ=0 is an eigenvalue.

If λλ is an eigenvalue of a matrix A, then 1/λ1/λ is an eigenvalue of A−1A−1.

Hence, ±i±i is the answer.

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  • 1
    $\begingroup$ Just looking at this answer gives me the feeling that I drank too much. But then "0 down vote" should have been "$0\!0$ down vote", but it is not. $\endgroup$ – Marc van Leeuwen Apr 28 '17 at 3:48

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