# Thread: Basic linear transformation question

1. ## Basic linear transformation question

Supposed T:U->U.

If $T \circ T = 0$, what does this mean? How could I use this in relation to rank or nullity of the linear transformation?

Thanks!

2. Well, if you think about the columns of T as individual vectors then you see each Matrix Vector Product results in the zero vector. So, that means the columns of T are in the kernel (null space) of T. Now what does the rank-nullity theorem say?

3. Originally Posted by lvleph
Well, if you think about the columns of T as individual vectors then you see each Matrix Vector Product results in the zero vector. So, that means the columns of T are in the kernel (null space) of T. Now what does the rank-nullity theorem say?
I am confused by what you means by 'the columns of T as individual vectors then you see each Matrix Vector Product' .

We have not covered matrixes all that much. I am more just interested in what happens to the rank and nullity of T. If it helps, U is finite dimensional.
I just am interested in what what $T \circ T = 0$ implies. Does it mean that it is surjective, injective, bijective, invertible, etc.

If that makes no sense, please excuse my English and ask me a question.

4. When one multiplies matrices they are multiply each row by each column. So if the matrix-matrix multiplication results in a 0 matrix, each column of T must have been in the kernel (null space) of T. This would imply that the matrix does not have {0} kernel and therefore is not invertible.

5. Originally Posted by lvleph
When one multiplies matrices they are multiply each row by each column. So if the matrix-matrix multiplication results in a 0 matrix, each column of T must have been in the kernel (null space) of T. This would imply that the matrix does not have {0} kernel and therefore is not invertible.
So we are saying if we have T composed of T equalling 0, then it is multiplying all the rows of T by all the columns of T of the matrix. The $columns of T \in ker(T)$. So what is the nullity and the rank then, and why? obviously I am still a little confused but am trying to think through this very hard.

6. Originally Posted by sabrepride
Supposed T:U->U.

If $T \circ T = 0$, what does this mean? How could I use this in relation to rank or nullity of the linear transformation?

Thanks!
You get that for any $x \in U, ~ T(T(x)) = 0 \Rightarrow$ for any $v \in T(U), ~ T(v) = 0$, where $T(U) = \{T(u) : u \in U\}$.

But $T(U) = Im(T)$. What do you get now?

7. So ignoring multiplication all together, I thought we were just speaking in terms of linear algebra, sorry.

If $T\circ T = 0$ then the kernel is not empty which means the nullity is not zero, i.e., the size of the kernel is not zero. By the rank nullity theorem we see that rank must be less than the dimenision of $U$. Therefore, $T$ is not surjective and thus not invertible.

8. Defunkt:
How do you know $T(U) = Im(T)$? I understand the first line of your answer, but do not see its implications. the Kernal is $\forall v \in U, T(v)=0$, and the image is the set of all z such that $\forall u \in U, T(u)=z$ correct?

lvleph: how do I know the kernel is not empty? That helps though, because the objective of this exercise was that the nullity(T) >= 1/2 dim(U). I have no clue how to prove this though.

Thank you both for your help!

9. Originally Posted by sabrepride
Defunkt:
How do you know $T(U) = Im(T)$? I understand the first line of your answer, but do not see its implications. the Kernal is $\forall v \in U, T(v)=0$, and the image is the set of all z such that $\forall u \in U, T(u)=z$ correct?

lvleph: how do I know the kernel is not empty? That helps though, because the objective of this exercise was that the nullity(T) >= 1/2 dim(U). I have no clue how to prove this though.

Thank you both for your help!
How did you define the image of a transformation? It is usually defined, for $S:V \to W$ as $Im(S) = \{S(v) : v \in V\}$, so for example if $V = \{1,2,3,4,5\}$ then $Im(S) = \{S(1),S(2),S(3),S(4),S(5)\}$, ie. the set of images of S.

This gives us that $Im(T) \subset Ker(T)$. Can you see why? If so, can you see how to finish now?

Don't look at the solution until you've given it a real try. I'm only leaving it here since I won't be able to attend this for a while.

Spoiler:

We have that $Im(T) \subset Ker(T) \Rightarrow dim ~ Im(T) \leq dim ~ Ker(T)$.

Now, using the fact that $dim(U) = dim ~ ker(T) + dim ~ Im(T)$ and the above inequality, we get that:
$dim(U) = dim ~ Ker(T) + dim ~ Im(T) \leq dim ~ Ker(T) + dim ~ Ker(T) = 2 * dim ~ Ker(T)$

So we get that $dim(U) \leq 2*dim ~ Ker(T) \Rightarrow dim ~ Ker(T) \geq \frac{1}{2}dim(U)$

10. $T(T(\vec{x}))=0\rightarrow im(T)\in\ker(T)\rightarrow rank(T)\leq nullity(T)=n-rank(T)$

$\rightarrow rank(T)\leq \frac{n}{2}~\&~nullity(T)\geq \frac{n}{2}$

11. Originally Posted by math2009
$T(T(\vec{x}))=0\rightarrow im(T)\in\ker(T)\rightarrow rank(T)\leq nullity(T)=n-rank(T)$

$\rightarrow rank(T)\leq \frac{n}{2}~\&~nullity(T)\geq \frac{n}{2}$
Well, the point was sort of that he would reach the solution by himself, but thanks regardless...

12. Originally Posted by Defunkt
How did you define the image of a transformation? It is usually defined, for $S:V \to W$ as $Im(S) = \{S(v) : v \in V\}$, so for example if $V = \{1,2,3,4,5\}$ then $Im(S) = \{S(1),S(2),S(3),S(4),S(5)\}$, ie. the set of images of S.

This gives us that $Im(T) \subset Ker(T)$. Can you see why? If so, can you see how to finish now?

Don't look at the solution until you've given it a real try. I'm only leaving it here since I won't be able to attend this for a while.

Spoiler:

We have that $Im(T) \subset Ker(T) \Rightarrow dim ~ Im(T) \leq dim ~ Ker(T)$.

Now, using the fact that $dim(U) = dim ~ ker(T) + dim ~ Im(T)$ and the above inequality, we get that:
$dim(U) = dim ~ Ker(T) + dim ~ Im(T) \leq dim ~ Ker(T) + dim ~ Ker(T) = 2 * dim ~ Ker(T)$

So we get that $dim(U) \leq 2*dim ~ Ker(T) \Rightarrow dim ~ Ker(T) \geq \frac{1}{2}dim(U)$
I really am forcing myself not to look at your 'spoiler or the lower post after I started to read it. However, I am seeing how this would be done incorporating the Rank-Nullity Theorem, however I am still lacking the ability to see how we know that from $T \circ T = 0$ that the $Im(T) \subset ker(T)$. T(U) is equal to the Im(T), I can see that. but the kernel is then T(x)=0 for all x in U, how is the image a subset of that?

Thanks!!

13. Originally Posted by sabrepride
I really am forcing myself not to look at your 'spoiler or the lower post after I started to read it. However, I am seeing how this would be done incorporating the Rank-Nullity Theorem, however I am still lacking the ability to see how we know that from $T \circ T = 0$ that the $Im(T) \subset ker(T)$. T(U) is equal to the Im(T), I can see that. but the kernel is then T(x)=0 for all x in U, how is the image a subset of that?

Thanks!!
Well, we want to show that $Im(T) \subset Ker(T)$. Take $v \in Im(T)$. We want to show that T(v) = 0. But since $v \in Im(T)$, we know that there exists some $u \in U : T(u) = v \Rightarrow T(v) = T(T(u)) = 0$ and therefore $T(v) = 0 \Rightarrow v \in Ker(T) \Rightarrow Im T \subset Ker T$