We need to find the rank of the matrix:

\begin{bmatrix} 0&c_{1}&-b_{1}&a_{2}\\ -c_{1}&0&a_{1}&b_{2}\\ b_{1}&-a_{1}&0&c_{2}\\ -a_{2}&-b_{2}&-c_{2}&0 \end{bmatrix}

Where all $a_{i},b_{i},c_{i}$ are non zero.

The only thing I could observe was the given matrix is skew-symmetric, and the determinant is $(a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2})^2.$However I could not proceed any further than that. The rank is ofcourse = the number of non zero rows in the row echelon form of the matrix, but taking that route seems to be too tedious , and I believe the purpose of giving a skew symmetric matrix was to exploit the (skewed)symmetry somehow.

The solution given in my book is :

  • Rank=$4$,if $det(A) \neq 0$. (This is straightforward)
  • Rank=$2$ otherwise, since "all determinants of the third order have the value zero".


  • What does the statement in quotes mean, and how is it true?

  • What connection does it have to finding the rank?

  • $\begingroup$ It means that all the $3\times 3$ submatrices have determinant equal to $0$. This implies that the rank of the matrix is strictly smaller than 3. On the other hand it is at least 2, since there is at least a $2\times 2$ submatrix with non zero determinant. $\endgroup$
    – Desperado
    Feb 9, 2021 at 9:45
  • $\begingroup$ @Desperado I see, but then How did we conclude all the $3x3$ submatrices are singular? Is there some kind of argument instead of manually considering all the submatrices(which are going to be many ) ? $\endgroup$
    – satan 29
    Feb 9, 2021 at 9:49
  • $\begingroup$ I think you are right to be puzzled about why the $3\times 3$ submatrices are singular: it only happens when $\det A=0$, so it is not easy to check. However what is clear is that every principal $3\times 3$ minor is $0$. Given that $\text{tr} A=0$ and $\det A=0$ in this case we have that the characteristic polynomial of $A$ is $X^4-0x^3 +pX^2-0X+0$ so that in this case $0$ is a double root and therefore the rank is at most $2$. But I still can't see directly why all $3\times 3$ minors vanish. $\endgroup$ Feb 9, 2021 at 20:26

2 Answers 2


An argument which avoids to consider all the $3\times 3$ submatrices is the following.

Consider the "twin" matrix $$A'=\begin{pmatrix} 0&c_{2}&-b_{2}&a_{1}\\ -c_{2}&0&a_{2}&b_{1}\\ b_{2}&-a_{2}&0&c_{1}\\ -a_{1}&-b_{1}&-c_{1}&0 \end{pmatrix}$$ If your matrix $A$ has zero determinant, then all the columns of $A'$ are in the kernel $\ker A$. By the Rank–nullity theorem it holds $rg(A)=4-\ker A\leq 4-rg(A')$, but $rg(A')\geq 2$, since $A'$ is at least a 2×2 submatrix with non zero determinant, hence $rg(A)\leq 2$. By the same reason $rg(A)\geq 2$.


There's basically two ways to address why a skew symmetric matrix $A$ never has odd rank.

1.) The most general uses bilinear forms and holds over any field (with extra careful definition of skewness in characteristic 2). This seems against the spirit of "connecting the rank with the determinant of a skew symmetric matrix" and the details are outside the scope.

2.) Here is a proof that hold over any field $\mathbb F$ with $\text{char}\big(\mathbb F\big) \neq 2$.
Suppose $\text{rank}\big(A\big)=r$ and $r\%2=1$ -- i.e. $r$ is odd.
Then $A$ has an $r\times r$ principal submatrix, $C_{r\times r}$ that is invertible-- i.e. $ \det\big( C_{r\times r} \big)\neq 0$
proof: Prove the existence of a principal submatrix of order $r$ in $M\in\Bbb F^{n\times n}, M=-M^T,\ \operatorname{rank}(M)=r$

But since $C_{r\times r}$ is a principal matrix, it is skew symmetric, which tells us
$\det\big(C_{r\times r}\big)=\det\big(C_{r\times r}^T\big)=\det\big(-I_r C_{r\times r} \big)=(-1)^r\cdot \det\big( C_{r\times r} \big)=-1\cdot \det\big( C_{r\times r} \big)$

$\implies 0 = 2\cdot \det\big( C_{r\times r} \big)$ $\implies 0 = \det\big( C_{r\times r} \big)\neq 0$
which is a contradiction

approach (2) actually works when $\text{char}\big(\mathbb F\big) = 2$ with a little bit of cleverness. First prove the lemma, working over the ring $\mathbb Z[\mathbf x]$ that for any $r\times r$ skew symmetric matrix matrix $S$ with zeros on the diagonal, and off diagonal elements are $x_{i,j} \in \mathbb Z[\mathbf x]$ then $\det\big(S\big) = 0$. One can do this by considering an arbitrary substitution homomorphisms of $x_{i,j}\to a_{i,j}\in \mathbb R$, and the determinant is zero for every possible substitution by the above argument-- i.e. in reals we see it implies 2 times the determinant is 0 hence the determinant is 0. By principle of permanence of identities the polynomial given by taking the $\det\big(S\big)\in \mathbb Z[\mathbf x]$ is equal to $ 0$. Now apply a different substitution homomorphism mapping the $x_{i,j}$ to elements in the desired field $\mathbb F$ of characteristic 2, and said polynomial is still zero -- i.e. we've proven that for odd r, any $r\times r$ skew symmetric matrix is singular (with explicit constraint of zeros on diagonal, which is the right definition of skew symmetry in char 2, and is redundant in other characteristics). From here the original argument (2) applies showing such a matrix cannot have odd rank.


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