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I have two questions about skew symmetric matrices.

  1. Why any skew symmetric matrix with $n$ (odd) number of rows and columns has $0$ as its only real eigenvalue.
  2. Is there a non zero skew symmetric matrix say $A$ such that $A + I$, where $I$ is identity matrix, gives orthogonal matrix.
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    $\begingroup$ The second is clearly impossible since any row/column with a non-zero off-diagonal entry has norm $\gt1$. $\endgroup$ – amd Mar 31 '17 at 21:21
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The defining equation $A = -A^T$ implies that $\det(A) = \det(-A^T) = (-1)^n \det(A^T) = (-1)^n \det(A)$ so if $n$ is odd we have $\det(A) = 0$ and zero is an eigenvalue of $A$. If $\lambda$ is an eigenvalue of $A$ with eigenvector $v \neq 0$, then

$$ \lambda (v^T v) = v^T Av = v^T (-A^Tv) = -(Av)^T v = - \lambda v^T v $$

so $\lambda = 0$.

Finally, if $(A + I)$ is orthogonal then

$$ I = (A + I)^T(A + I) = (A^T + I)(A + I) = A^T A + A^T + A + I = A^T A + I$$

so $A^T A = 0$ which implies that $A = 0$.

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