# Thread: Nullity of a transformation

1. ## Nullity of a transformation

T(x,y) = ((2x-(y/2)), -x+3y, 2y)

I have found the kernel which is 0 how do we find the nullity of this?

2. Originally Posted by adam_leeds
T(x,y) = ((2x-(y/2)), -x+3y, 2y)

I have found the kernel which is 0 how do we find the nullity of this?
nullity is the dimension of kernel(T),..

so, dim(ker(T)) = . . .

3. Do you know what the dimension of the "trivial" subspace, the one containing only the 0 vector, is?

4. Originally Posted by dedust
nullity is the dimension of kernel(T),..

so, dim(ker(T)) = . . .
0?

5. Originally Posted by adam_leeds
0?
yes,..

6. Originally Posted by dedust
yes,..
if the kernel was x = 2 y = 0, would the nullity be 1?

7. Originally Posted by adam_leeds
if the kernel was x = 2 y = 0, would the nullity be 0?
you mean the kernel is $\left({\begin{matrix}
2\\
0\\
\end{matrix}}\right)$
?
what is the dimension of the space that spanned by this vector?

8. Originally Posted by dedust
you mean the kernel is $\left({\begin{matrix}
2\\
0\\
\end{matrix}}\right)$
?
what is the dimension of the space that spanned by this vector?
Sorry ill re phrase

to find the kernel of a transformation

you make the equations = 0

and solve for x and y

if 1 of x and y doesnt = 0

is the dimension 1?

and if they are both non zero solutions then the dimension is 2?

9. Originally Posted by adam_leeds
if the kernel was x = 2 y = 0, would the nullity be 1?
Run the vector (2,0) through your transformation. What do you get?

10. Originally Posted by ands!
run the vector (2,0) through your transformation. What do you get?
4, -2, 0

11. Originally Posted by adam_leeds
4, -2, 0
Therefore it is NOT in the kernel of the linear transformation.

But, since you said "I have found the kernel which is 0" in your first post, I think you were asking in general, not for this particular transformation. It is not possible for the kernel of a linear transformation to be a single non-zero vector!

The kernel of a linear transformation, A, is the set of all vectors, v, such that Av= 0. That is not true for (2, 0). Further, if the kernel is not "trivial", that is, if Av= 0 for any non-zero v, it cannot consist of a single vector! If v is a non-zero vector such that Av= 0 then A(xv)= xAv= 0 for any number x so there must exist an infinite number of vectors in the kernel. The kernel of a linear transformation is a subspace, possibly trivial, which is why it has a dimension.