Simple basis / spanning set question (T / F)

[theRukus]

Homework Statement

Let T: R⁴ --> R⁷ be a linear map whose kernel has the basis of v = [1 0 1 2]^T. What is the dimension of the image of T?

The Attempt at a Solution

I have a very loose understanding of kernel and image, and am trying very hard to get this question. From my understanding, the kernel of T (the one in this problem) is the set of all vectors X in R⁴ that satisfy TX = 0. The image of T is the set of all vectors Y in R⁷ that satisfy Y = TX.So, from that understanding (which is very little):

Since the basis for the kernel only has one vector, X = [1 0 1 2]^T (or any multiple of it). So.. The kernel has a dimension of 1.. (correct?) I'm completely unsure as to where to go from here..

Any help is greatly appreciated. Thank you.

** Sorry, I just wanted to note that I know I titled the post wrong. I was going to post a different question, but figured it out..

 

[Coto]

The theorem you're looking to use/understand is this one:

http://en.wikipedia.org/wiki/Rank-nullity_theorem

(You are correct about the kernel having one dimension.)