Let T: R3 --> R3 be the linear transformation that projects u onto v = (3,0,4)
Find the rank and nullity of T
So let u=(x,y,z)
The Attempt at a Solution
So I know that
T(u) = proj. u onto v
T(u) = [(3x + 4z)/ 25](3,0,4)
T(u) = 0
This is where I get confused, in a simalar example from my text (which skips some steps) leads me to belive that I can set 3x + 4z = 0.
Is not the nullity(T) = dim(ker(T))?
So would'nt I have to
T(u) = [(3x + 4z)/ 25](3,0,4) = (0,0,0)
Create an augmented matrix and find the nullspace?
I get a different answer than the book is getting?