# 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?

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?

0?
yes,..

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

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?

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