1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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 $\det(A) = \det(A^T) = \det(-A)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.