Permutations Matrices Question

**Aquameatwad** · Sep 23rd 2010, 04:56 PM

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.

**Plato** · Sep 23rd 2010, 05:20 PM

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.

**Aquameatwad** · Sep 23rd 2010, 05:56 PM

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

**HallsofIvy** · Sep 24th 2010, 05:08 AM

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

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

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

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

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