How does the rank nullity theorem apply in this case?

Question

Imagine a straight line through the origin in $\mathbb{R}^4$. It is a vector space of dimension 1. Now let this undergo a linear tranformation through a 4$\times $4 invertible matrix. Rank of the matrix is 4. Dimension of the kernel is at least 0. They both sum to a number greater than or equal to 4. However the dimension of the departure space is only 1. Where am I going wrong?

Answer 1

When you say the rank is 4 that is with reference to the linear transformation on $\mathbb R^4$ and not on the line.

Answer 2

A linear map (or any other function) needs to have its domain and codomain specified. The matrix represents a map $M: \mathbb{R}^4 \rightarrow \mathbb{R}^4$ which does indeed have rank 4, but by restricting it to the line we are now considering a different linear map $M': L \rightarrow \mathbb{R}^4$. This linear map has rank 1 (It's image - which is the same as the image of $L$ under $M$ - is another line) and Rank-nullity works out correctly.