0

I know that $x^TAy = y^TAx$ is true for symmetric quadratic matrices, but, it is true for non symmetric quadratic matrices?

Widawensen
  • 8,172
beta developer
  • 3
  • 6
  • 3
    Try a simple $2\times 2$ example. It's not hard to find a counter-example – Winther Mar 06 '18 at 15:19
  • No it's not true, it might not even make sense to write that if $A$ isn't square to begin with. – Jürgen Sukumaran Mar 06 '18 at 15:19
  • 3
    Assuming these are real matrices, we always have $x^T A y = (x^T Ay)^T = y^T A^T x$. Therefore, if $x^TAy = y^T A x$, we have $y^TAx = y^TA^Tx$. Since this must hold for all vectors $x$ and $y$, we can conclude that $A=A^T$, so symmetry is necessary. –  Mar 06 '18 at 15:21
  • Duplicate : https://math.stackexchange.com/questions/5063/how-do-i-prove-that-xtay-ytax-if-a-is-symmetric?rq=1 – Exodd Mar 06 '18 at 15:28
  • 1
    Any non symmetric matrix will be a counterexample. – copper.hat Mar 06 '18 at 15:40

1 Answers1

1

No. Let $e_i$ denotes the vector with a $1$ at the $i$-th position and zeros elsewhere. If $A$ is not symmetric, then $a_{ij}\ne a_{ji}$ for some $i\ne j$, but then $e_i^TAe_j=a_{ij}\ne a_{ji}=e_j^TAe_i$.

user1551
  • 139,064