If $A$ is a symmetric $n × n$ matrix and $B$ is a skew symmetric $n × n$ matrix, which of the following are true?

(a) $ABA$ is symmetric

(b) $ABA$ is skew-symmetric

(c) $AB^2A$ is symmetric

(d) $AB^2A$ is skew-symmetric

I know that b and d holds true.

I am unsure of A and C

However, for a, how does multiplying ABA preserve symmetry, but squaring B preserves symmetry as well?

  • $\begingroup$ Do you know that $(AB)^t=B^tA^t$? $\endgroup$ – Javi Jul 22 '18 at 10:59
  • $\begingroup$ @Javi right. But I don't understand how the composition of matrices can effect whether or not it is symmetric or skew-symmetric $\endgroup$ – FireMeUP Jul 22 '18 at 10:59
  • $\begingroup$ I'll post an answer $\endgroup$ – Javi Jul 22 '18 at 11:02
  • $\begingroup$ @Javi thank you for your clarifications $\endgroup$ – FireMeUP Jul 22 '18 at 11:02
  • $\begingroup$ @FireMeUP: Use this MathJax for your future posts! $\endgroup$ – Chinnapparaj R Jul 22 '18 at 11:11

Use the property $(AB)^t=B^tA^t$ to compute the transpose of each matrix and the fact that $A$ is symmetric and $B$ is skew-symmetric. For example, $(ABA)^t=A^tB^tA^t=A(-B)A=-(ABA)$. Then $ABA$ is skew-symmetric (and not symmetric in general).

  • $\begingroup$ how does this change for B^2. Do we have to compute B^2t? $\endgroup$ – FireMeUP Jul 22 '18 at 11:03
  • $\begingroup$ Yes, but that's just $B^tB^t=(B^t)^2$. $\endgroup$ – Javi Jul 22 '18 at 11:05
  • $\begingroup$ ahhh I I see. So in the case for B^2. I have that AB^2A is symmetric, but not skew symmetric since using the transpose condition you have give me, we reduce back to (A^TBB*A^T) which cannot be negative $\endgroup$ – FireMeUP Jul 22 '18 at 11:08
  • $\begingroup$ Yeah, that's it :) $\endgroup$ – Javi Jul 22 '18 at 11:08
  • 1
    $\begingroup$ thank you for your help $\endgroup$ – FireMeUP Jul 22 '18 at 11:11

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