Let B be any real, nonsingular $nxn$ matrix, where n is even, and set $A=B-B^T$. Show that A does not admit an LU decomposition without pivoting.

I know that A is a skew symmetric matrix. The 2x2 case is

0 -a

a 0

It is easy to see we can't eliminate a so we must pivot. What I don't understand is why n must be even? I think even odd cases show this... Thanks.

  • $\begingroup$ You should consider the eigenvalues for odd skew-symmetric matrices. $\endgroup$ Oct 21 '16 at 2:24
  • $\begingroup$ Is it because they are imaginary? So the matrix isn't real? $\endgroup$
    – MathIsHard
    Oct 21 '16 at 2:31
  • 1
    $\begingroup$ If $n$ is odd then you have a real eigenvalue. $\endgroup$ Oct 21 '16 at 2:32

I don't think any invertible skew-symmetric real matrix admits an LU decomposition without pivoting.

From $A^T=-A$, if $A=LU$ we get $$LU=-(LU)^T=-U^TL^T.$$ We are assuming that $A$ is invertible, so $L$ and $U$ are also invertible. Thus $$(U^T)^{-1}L=-L^TU^{-1}.$$ Here the left-hand-side is lower triangular, and the right-hand-side is upper triangular; this implies that both are diagonal. So $(U^T)^{-1}L=D$ for an invertible diagonal matrix $D$. We can write $L=U^TD$. We also have $-L^TU^{-1}=D$, and we obtain $$ L^T=-DU, $$ which we can write as $L=-U^TD.$ It follows that $U^TD=-U^TD$, which would imply that $U^TD=0$, a contradiction since $U$ and $D$ are invertible.


If $n$ is odd then the characteristic polynomial for $A$ emits at least one real root. And zero happens to be a root which means $A$ is singular.

As an exercise, try to see if you can prove that $A$ is singular when $n$ is odd.

Hint: Consider $\det(A) = \det(A^T) = \det(-A)$.


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