# Linear algebra proof with operator T^2 = cT

1. Jun 17, 2009

### evilpostingmong

1. The problem statement, all variables and given/known data
Let T:V--->V be an operator satisfying T^2=cT c=/=0.
Show that V=U$$\oplus$$kerT U={u l T(u)=cu}

2. Relevant equations

3. The attempt at a solution
Now before I start, just one quick question about ker T:
U seems to be an eigenspace since T(u)=cu with c the eigenvalue.
But that must mean that the kernel has 0 as the only element since
0 is not an eigenvector so it can't be in the eigenspace, right?
And, since T^2=cT, c cannot be 0 otherwise we would have T(0).

2. Jun 17, 2009

### tiny-tim

Hi evilpostingmong!

(have a not-equal: ≠ and try using the X2 tag just above the Reply box )
T2 = cT means T is a projection …

so suppose eg V is R3, and T is the projection onto the x-y plane (with c = 1),

then kerT is the z-axis.