0

Let T:V→W be a linear transformation where V is an infinite-dimensional vector space and W is a finite-dimensional vector space. What is the dimension of the kernel?

Hello everyone. Sorry my English, it's not my first language.

I tried doing this:

T(v1)=w1

T(v2)=w2

...

T(vn)=wn

But V has infinite vectors, so remaining vectors are in the kernel.

I don't know if I can use dim V = dim KerT + dim ImT here.

I'm not sure of the things I said. If someone may help me I would appreciate a lot. Thank you very much for attention!

what i'm trying to say

jgon
  • 28,469
Harry
  • 13
  • 4

1 Answers1

3

Suppose the dimension of the kernel is finite, so $\ker f$ has $\{y_1,\dots,y_n\}$ as basis.

If $\{f(x_1),\dots,f(x_m)\}$ is a basis of the image of $f$, prove that $$ \{x_1,\dots,x_m,y_1,\dots,y_n\} $$ is a spanning set for $V$ (actually a basis).

egreg
  • 238,574
  • May I use this: https://math.stackexchange.com/questions/3070298/how-can-i-prove-this-property-in-a-linear-transformation?noredirect=1#comment6330417_3070298? – Harry Jan 11 '19 at 23:26
  • If we have a linear transformation T:V→W and the dimension of V is greater than the dimension of W, then the kernel of T has to have a dimension at least as large as the difference. So, when V is infinite dimensional and W has a finite dimension, then the kernel has to be infinite dimensional. – Harry Jan 11 '19 at 23:26
  • My question is, can I use the rank-nullity theorem in this case? When I have an infinite dimensional vector space? – Harry Jan 11 '19 at 23:29
  • @Harry I didn't use the rank-nullity theorem. What I proved (in a way very similar to the theorem) that if the image is finite dimensional and the kernel is finite dimensional, then the domain is finite dimensional as well. Hence in your case the dimension of the kernel must be infinite. – egreg Jan 11 '19 at 23:45
  • I understood! Thank you very much for your help. I'm very happy now. – Harry Jan 11 '19 at 23:51