Simple basis / spanning set question (T / F)

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The discussion centers on a linear map T: R4 --> R7 with a kernel basis represented by the vector v = [1 0 1 2]T. The kernel's dimension is confirmed to be 1, indicating that there is one degree of freedom in the solution space of the equation TX = 0. According to the Rank-Nullity Theorem, the dimension of the image of T can be determined by subtracting the kernel's dimension from the domain's dimension, leading to the conclusion that the image's dimension is 3.

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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..
 
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