Question

show that rank of a skew symetric matrix cannot be one​

Answer 1

For a skew symmetric (real) matrix, the eigenvalues are all purely imaginary. This is because if Av=λvAv=λv, then we have λ⟨v,v⟩=⟨λv,v⟩=⟨Av,v⟩=⟨v,−Av⟩=⟨v,−λv⟩=−λ¯¯¯⟨v,v⟩λ⟨v,v⟩=⟨λv,v⟩=⟨Av,v⟩=⟨v,−Av⟩=⟨v,−λv⟩=−λ¯⟨v,v⟩, so we conclude that λ=−λ¯¯¯λ=−λ¯, i.e., that λλis purely imaginary. Here, we're using an Hermitian inner product.

For a real matrix, complex eigenvalues come in conjugate pairs, so the rank must be even.

Is there a succinct proof for the fact that the rank of a non-zero skew-symmetric matrix (A=−ATA=−AT) is at least 2? I can think of a proof by contradiction: Assume rank is 1. Then you express all other rows as multiple of the first row. Using skew-symmetric property, this matrix has to be a zero matrix.