Nullity of Linear Transformation T:M_2x3(F) -> M_2x2(F): 4

[_F_]

Homework Statement

Find the nullity of the linear transformation T:M_2x3(F) -> M_2x2(F) defined by:

T([a11, a12, a13; a21, a22, a23]) = ([2*a11 - a12, a13 + 2*a12; 0, 0])

The Attempt at a Solution

N(T) = {x in M_2x3(F) | T(x) = 0}

2*a11 - a12 = 0 => a11 = (a12)/2
a13 + 2*12 = 0 a13 = -2*a12

N(T) = {[a/2, a, -2*a; 0, 0, 0] | a in F}

Basis(N(T)) = {[1/2 1 -2; 0, 0, 0]}

nullity(T) = 6

But nullity(T) should be 4...

Any help is appreciated.

 

[Defennder]

N(T) = {[a/2, a, -2*a; 0, 0, 0] | a in F}
Basis(N(T)) = {[1/2 1 -2; 0, 0, 0]}
nullity(T) = 6
But nullity(T) should be 4...

Why are these entries 0's? Look at the formula given for T. Do a₂₁,a₂₂,a₂₃ appear anywhere in the transformed 2x2 matrix?

 

[_F_]

Why are these entries 0's? Look at the formula given for T. Do a₂₁,a₂₂,a₂₃ appear anywhere in the transformed 2x2 matrix?

You're right. But that said, if you take away the zero's you still get a 1x3 matrix whose dimension is 3...

 

[Defennder]

You meant I a 2x3 matrix I presume. And it doesn't make sense to talk about the dimension of a matrix. You meant the dimension of the nullspace. And it isn't 3. How did you arrive at that?