I know that the rank of a skew-symmetric matrix is even. I just need to find a published proof for it. Could anyone direct me to a source that could help me?
The rank of skew-symmetric matrix is even
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0Maybe this https://math.stackexchange.com/questions/1555045/show-that-rank-of-skew-symetric-is-even-number – 2017-07-04
3 Answers
You can see Hoffman's book on linear algebra last chapter on Bilinear forms which says rank of skew symmetric matrix is always even.
Notice that over a field of characteristic 2, the matrix $\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$ is skew-symmetric, and has rank $3$. So I think you are really looking for a proof that an alternating matrix (skew-symmetric with zeros on the diagonal) has even rank.
The proof for this usually comes up in looking at the theory of bilinear forms, there are a few references mentioned in the comments that contain proofs, but another option is in "Classical Groups and Geometric Algebra" by Larry Grove.
An alternate proof to this :
Skew Symmetric matrix follows $A = -A^T$.
Now, Suppose there exists at least one non zero quantity "a" such that a largest non singular sub matrix exists :
\begin{pmatrix} 0 & a \\ -a & 0 \\ \end{pmatrix} The order of the largest non singular sub matrix in a given matrix is defined as the rank of the matrix . Clearly it's rank = 2. Now, you can generalise for presence of more non zero entities in the matrix.
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5This does not make any sense, as far as I can tell. – 2017-07-03