Thread: Permutations Matrices Question

This is a homework problem i had. Help!

Explain why the dot product of x and y equals the dot product of Px and Py.
Then From (Px)^T(Py)=x^Ty deduce that P^TP=I for any permutation. With x=(1,2,3) and y=(1,4,20) choose P to show that Pxdoty is not always xdotPy.

I dare say that you have most of us at a disadvantage.
I for one have absolutely no idea what that question means.
You must define the terms and explain the notation.

I think its referring to the Identity Matrix I. and the different permutations which are P. And ^T means to find the dot product of

P is a "permutation" matrix: every row and column contains exactly one "1" and all other entries are 0. For example,
$\displaystyle Px= \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}y \\ z \\ x\end{bmatrix}$
just permutes the entries x, y, and z.

If, instead of $\displaystyle x= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$ we use $\displaystyle v= \begin{bmatrix}u \\ v \\ w\end{bmatrix}$ then $\displaystyle Pv= \begin{bmatrix}v \\ w \\ u\end{bmatrix}$.

$\displaystyle (Px)^TPv= \begin{bmatrix}y & z & x\end{bmatrix}\begin{bmatrix}v \\ w \\ u\end{bmatrix}= vw+ wz+ ux= \begin{bmatrix}x & y & z\end{bmatrix}\begin{bmatrix}u \\ v \\ w\end{bmatrix}= x^Tv$
Essentially since both x and v are permuted in the same way, the "corresponding terms" still match up.

However,
$\displaystyle (Px)^Tv= \begin{bmatrix}y & z & x\end{bmatrix}\begin{bmatrix}u \\ v \\ w\end{bmatrix}= uy+ vz+ wx$
while
$\displaystyle x^T(Pv)= \begin{bmatrix} x & y & z\end{bmatrix}\begin{bmatrix}v \\ w \\ u\end{bmatrix}= vx+ wy+ uz$
so $\displaystyle (Px)^Tv$ is not, in general, equal to $\displaystyle x^T(Pv)$.