Suppose TT is a linear map from F4\mathbf F^4 to F2\mathbf F^2 such that

null T={(x1,x2,x3,x4)∈F4:x1=5x2 and x3=7x4}\text{null } T = \{(x_1,x_2,x_3,x_4) \in \mathbf F^4 : x_1 = 5x_2 \text{ and } x_3 = 7x_4\}

Prove that TT is surjective.

By 3.22

dim⁡range T+dim⁡null T=4\dim \text{range }T + \dim \text{null }T = 4

Since dim⁡null T=2\dim \text{null }T = 2 (exercise to the reader) this implies dim⁡range T=2\dim \text{range }T = 2 so we can let v1,v2v_1,v_2 be a basis of range T\text{range }T, extending (2.33) this basis to a basis of F2\mathbf F^2 doesn't add anything, implying v1,v2v_1,v_2 is already a basis for F2\mathbf F^2.

I feel there should be a theoremin the book I could cite to show if UU is a subspace of VV, U=VU = V iff dim⁡U=dim⁡V\dim U = \dim V.

Proof: Let u1,…,umu_1,\dots,u_m be a basis of UU and extend it to a basis of VV. Since dim⁡U=dim⁡V\dim U = \dim V extending it does nothing, thus u1,…,umu_1,\dots,u_m is already a basis for VV and so U=VU = V.