What is the Dimension of the Intersection of Two Kernels in a Vector Space?

[daniel_i_l]

Homework Statement

Given transformations T_1, T_2:V->F where V is a vector space with the dimension n over the field F, T_1 , T_2 =/= 0. If N_1 = KerT_1 , N_2 = KerT_2 and N_1 =/= N_2 find dim(N_1 intersection N_2)

Homework Equations

dim(A+B) = dimA + dimB - dim(A intersection B)

The Attempt at a Solution

First of all, dimImT_1 = dimImT_2 = 1 so N_1 = N_2 = n-1.
Also, N_1 + N_2 in V so
dim(N_1 + N_2) = n-1+n-1-dim(N_1 intersection N_2)
<= n and so we get that dim(N_1 intersection N_2) >= n-2.
But since N_1 intersection N_2 in N_1 we get
n-1 >= dim(N_1 intersection N_2) >= n-2. But since N_1 =/= N_2 obviously
dim(N_1 intersection N_2) =/= n-1 and so dim(N_1 intersection N_2) =n-2.
Is that right? Did I leave out any important step?
Thanks.

 

[HallsofIvy]

Homework Statement

Given transformations T_1, T_2:V->F where V is a vector space with the dimension n over the field F, T_1 , T_2 =/= 0. If N_1 = KerT_1 , N_2 = KerT_2 and N_1 =/= N_2 find dim(N_1 intersection N_2)

Homework Equations

dim(A+B) = dimA + dimB - dim(A intersection B)

The Attempt at a Solution

First of all, dimImT_1 = dimImT_2 = 1 so N_1 = N_2 = n-1.

Why is that true? What if T_1 were the identity transformation? How do you conclude that the dimension of the image of an arbitrary non-trivial linear transformation is 1 no matter what n is?

Also, N_1 + N_2 in V so
dim(N_1 + N_2) = n-1+n-1-dim(N_1 intersection N_2)
<= n and so we get that dim(N_1 intersection N_2) >= n-2.
But since N_1 intersection N_2 in N_1 we get
n-1 >= dim(N_1 intersection N_2) >= n-2. But since N_1 =/= N_2 obviously
dim(N_1 intersection N_2) =/= n-1 and so dim(N_1 intersection N_2) =n-2.
Is that right? Did I leave out any important step?
Thanks.

Think about this special case: V= R³, T_1(x,y,z)= (x,y,z) and T_2= (x, y, 0). What are N_1 and N_2?

 

[daniel_i_l]

Sorry, I meant that both T_1 and T_2 are from V to F:
T_1:V->F
T_2:V->F
Thanks.