Question

I need help to understand the concept, please
\[T:R^2\rightarrow R^3~ ~ where R^3 \]is range of T, and it is said in a matrix A2x3 \(T: R^3 \rightarrow R^2\)
I am confused.

Answer 1

What part don't you understand?

Answer 2

for theorem : The linear transformation T: V--> W is
1/ one to one iff ker(T) =0
2/ onto iff Rng(T) =W
so, we calculate base on the range, not the domain? while calculating ker(T), we have to use V??

Answer 3

V is preimage set, and W is image one, right? in a matrix 2x3 how a column space is image of row space?

Answer 4

Slow down. You're posting different things. What exactly is your question?

Answer 5

see my second post, it's a theorem which I am confused

Answer 6

Well for the first part, it's true because:
Say d spans ker(T), then multiple values of d can yield 0 when the transformation is applied. Meaning T(span (ker(T))) = 0. Since the function is one-to-one, all span ker(T) must map from one point, hence all span ker(T) must be one point. This is only possible with the zero vector because span (0) can only be 0. Hence span ker(T) = ker (T) = 0.

Answer 7

Got it?

Answer 8

It seems not relate to my question :) I ask about range and domain of T

Answer 9

You only gave me a theorem. You didn't give any information about T.

Answer 10

to any matrix transformation T: R^n--> R^m with mxn matrix A has a natural subspaces, T : V ---> W
ker ( T) = null space of A (subspace of R ^n)
Rng (T) = column space of A (subspace of R ^m)
and dim (ker (T) + dim (Rng (T) = dim ( V)
you see, ker (T ) is calculate based on \(\huge R^n\) while Rng(T) is based on \(\huge R^m\)
and at the end up, sum of dimensions is based on \(\huge V\) which is \(\huge R^n\)
why??

Answer 11

Your question is "why sum them"?

Answer 12

I am confused when Rng (T) is based on R^m , how it relates to R^n which is another subspace of A

Answer 13

Okay. Well, If T(V) can be written as BV then span( BV )= Range (T).
Do you agree with this?

Answer 14

ok

Answer 15

Then BV = W.
So, BV-W = 0.
For the Ker(T), then BV-0=0
BV = 0.
V must be must be in the nullspace of B. The definition of nullspace is whatever makes the equation equal to 0.
In your case, A was used instead of B, so ker(T) must be in the nullspace of A.

Answer 16

Got you, thanks a ton.!!:)

Answer 17

You're welcome.

Answer 18

wonder why I am always confused the concepts :(

Answer 19

Well, it takes time. Just try to apply other things you learned and you'll be fine.

Answer 20

thank you