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

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

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You should consider the eigenvalues for odd skew-symmetric matrices.

– Jacky Chong

Oct 21 at 2:24

Is it because they are imaginary? So the matrix isn’t real?

– ryBear

Oct 21 at 2:31

1

If nn is odd then you have a real eigenvalue.

– Jacky Chong

Oct 21 at 2:32

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2 Answers

2

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

From AT=−AA^T=-A, if A=LUA=LU we get LU=−(LU)T=−UTLT.LU=-(LU)^T=-U^TL^T. We are assuming that AA is invertible, so LL and UU are also invertible. Thus (UT)−1L=−LTU−1.(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 (UT)−1L=D(U^T)^{-1}L=D for an invertible diagonal matrix DD. We can write L=UTDL=U^TD. We also have −LTU−1=D-L^TU^{-1}=D, and we obtain

LT=−DU,

L^T=-DU,

which we can write as L=−UTD.L=-U^TD. It follows that UTD=−UTDU^TD=-U^TD, which would imply that UTD=0U^TD=0, a contradiction since UU and DD are invertible.

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

As an exercise, try to see if you can prove that AA is singular when nn is odd.

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