# What is the dimension of the kernel of a linear transformation from infinite dimensional to finite dimensional?

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 is the point of writing down $T(v_k) = w_k$ where the $v_k,w_k$ are undefined??? – copper.hat Jan 9 '19 at 21:27
• Sorry about this, i tried to say that v belongs to V and w belong to W. – Harry Jan 9 '19 at 21:36
• That is still meaningless. – copper.hat Jan 9 '19 at 21:36
• I understand that. But what is the relevance to the question other than noise? Without specifying $v_k,w_k$ it means nothing. – copper.hat Jan 9 '19 at 21:44
• Look, I have a passing familiarity with algebra, that is not the issue. The issue is that unless you specify what $v_k, w_k$ are the statement $T(v_k) = w_k$ is nothing but a bunch of symbols. Nowhere in your question do you specify what the $v_k,w_k$ are. – copper.hat Jan 9 '19 at 22:00

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).