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I know it's possible to get a 3-dimensional subspace by using the skew-symmetric matrices, but is it possible to get anything larger than that? How would I go about proving there is no 4-dimensional subspace?

Edit: To clarify, by rank 2 variety I mean the variety of matrices with rank<=2, i.e. det(M)=0.

nekcihcyerewolf
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    What about matrices with zero first column? – Sasha Mar 01 '22 at 19:11
  • @Sasha Thanks, I don't know how I forgot about that. – nekcihcyerewolf Mar 01 '22 at 19:39
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    See this post. The maximal possible dimension in this case is $3^2 - 3 = 6$; Sasha's comment above is an example of such a subspace with this dimension. For $n \times n$ matrices, the largest subspace contained in the subspace of singular $n\times n$ matrices has dimension $n^2 - n$. – Ben Grossmann Mar 01 '22 at 20:56

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