1: Let V and W be vector spaces over a field F, and let T : V W be a linear transformation. Suppose that ker(T) and im(T) are both finite dimensional, that {v1,...vn} is a basis for ker(T), and that {w1,...wm} is a basis for im(T). For each i {1, . . . m}, consider the vector wi. Since wi is in the image of T, there is a vector ui in V with T(ui) = wi. Pick such a vector, for each i. Now think about the set of vectors {v1,...vn,u1,...um} in V. a) Show that this set of vectors is linearly independent. b) Let v be a vector in V. Show that there is some vector u in V, where u is a linear combination of the vectors u1, . . . um, such that T (v) = T (u). c) Explain why vu can be expressed as a linear combination of the vectors v1,...vn. d) Now explain why v can be expressed as a linear combination of the vec- tors v1,...vn,u1,...um. This shows that {v1,...vn,u1,...um} is a basis for V. Therefore: dim(V ) = n + m = dim(ker(T )) + dim(im(T )). This result is called the Rank-Nullity Theorem. So, good job proving the Rank- Nullity Theorem