Prove the following identity

**FGT12** · December 20th 2011, 09:36 AM

Show that for $A \in M_n(\mathbb{R})$ and $x,y \in \mathbb{R}^n$

$x \cdot Ay = y \cdot Ax$

I get
$y \cdot Ax = y^t Ax$ but i cannot see where to go now

I think it may involve $(AB)^t=B^tA^t$

**ILikeSerena** · December 20th 2011, 10:05 AM

Hi FGT12!

You're on the right track.
Can you say what $(y^tAx)^t$ is?

**Opalg** · December 20th 2011, 12:18 PM

Originally Posted by FGT12

Show that for $A \in M_n(\mathbb{R})$ and $x,y \in \mathbb{R}^n$

$x \cdot Ay = y \cdot Ax$

I get
$y \cdot Ax = y^t Ax$ but i cannot see where to go now

I think it may involve $(AB)^t=B^tA^t$