# Projection matrix questions

1. Mar 29, 2007

### EvLer

I am studying for exam and something does not make sense anymore:
looking at projection matrix, how come P=P2 where
P2 = A(ATA)-1ATA(ATA)-1AT
= A(ATA)-1AT
= P

but then they also say that cancelations (like distributing inverse operation and having AA-1 = I type things) are possible only if A is invertible, so does that mean that ATA is invertible?

EDIT: might as well ask this:
find matrix of projection p on plane x+y+2z = 0
my attempt: I know that the normal vector is (1, 1, 2) and then... not sure where to go with that....

Last edited: Mar 29, 2007
2. Mar 30, 2007

### HallsofIvy

Staff Emeritus
Yes, although a projection map is not, in general, invertible, it's not too difficult to show that PTP always is.

Yes, (1, 1, 2) is normal to the given plane and, since that has length $\sqrt{6}$, $(1/\sqrt{6},1/\sqrt{6},2/\sqrt{6})$ is a unit vector normal to the plane.

The standard way to find the matrix for a linear transformation (in a given basis) is to see what it does to each of the basis vectors. Since the dot product of (1, 0, 0) with $(1/\sqrt{6},1/\sqrt{6},2/\sqrt{6})$ is $1/\sqrt{6}$, its projection onto that vector is (1/6, 1/6, 1/3) and so its projection onto the plane is (1, 0, 0)- (1/6, 1/6, 1/3)= (5/6, -1/6, -1/3). Similarly, the projection of (0, 1, 0) onto the plane is (-1/6, 5/6, -1/3) and the projection of (0, 0, 1) onto the plane is (-1/3, -1/3, 2/3). The matrix representation of the projection is the matrix having those vectors as columns (in that order).

3. Mar 30, 2007

### EvLer

thanks I think I got it now :)