What is the nullity and geometric description of the kernel and range of T?

[trojansc82]

Homework Statement

Use the given information to find the nullity of T and give a geometric description of the kernel and range of T.

T is the projection onto the vector v = (1,2,2):

T(x,y,z) = (x + 2y + 2z)/9 (1,2,2)

Homework Equations

Kernel of T = T(v) = 0.

Nullity of T = dimension of the kernel of T

The Attempt at a Solution

I created an augmented matrix that looks like this:

[ 1/9 2/9 2/9 | 1
2/9 4/9 4/9 | 2
2/9 4/9 4/9 | 2 ]

Row reducing I get just one vector because the other two are a scalar.

However, the answer in the book says the nullity is 2, so I am off.

 

[HallsofIvy]

You are overworking this. Given a vector z, any vector v can be written as au+ bz, where u is perpendicular to z, so that T(v)= bz. In particular, the range of T is the subspace spanned by z and the kernel is the orthogonal complement to that. Here, the range is the one-dimensional subspace spanned by <1, 2, 2>. It's orthogonal complement, the kernel, is the two dimensional space of all vectors <x, y, z> such that <x, y, z>.<1, 2, 2>= x+ 2y+ 2z= 0. x= -2y- 2z so <x, y, z>= <-2y- 2z, y, z>= <-2y, y, 0>+ <-2z, 0, z>= y<-2, 1, 0>+ z<-2, 0, 1> so the kernel has basis {<-2, 1, 0>, <-2, 0, 1>}.