The solutions are ready. Please read them and ask if something is unclear. I wrote the proof for 2.1 in some other words, maybe it will be simpler for you.
2.1. A linear operator 𝐿: 𝑉 → 𝑊 is one-to-one if and only if 𝐾𝑒𝑟(𝐿) = 0𝑉 .
Proof. If and only if means two things:
1) If 𝐿 is one-to-one then 𝐾𝑒𝑟(𝐿) = 0𝑉 and 2) If 𝐾𝑒𝑟(𝐿) = 0𝑉 then 𝐿 is one-to-one.
Both will get proof by a contradiction.
1) Let 𝐿 be one-to-one but 𝐾𝑒𝑟(𝐿) ≠ 0𝑉 , i.e. there is a nonzero vector 𝑣0 ∈ 𝐾𝑒𝑟(𝐿), i.e.
𝐿(𝑣0 ) = 0𝑊 . But then 𝐿(0𝑉 ) = 𝐿(𝑥0 ) while 0𝑉 ≠ 𝑥0 ,
which is a contradiction with 𝐿 to be one-to-one, Q.E.D.
2) Let 𝐾𝑒𝑟(𝐿) = 0𝑉 but 𝐿 is not one-to-one. This means there are 𝑥1 ≠ 𝑥2 : 𝐿(𝑥1 ) = 𝐿(𝑥2 ).
But then 𝐿(𝑥1 − 𝑥2 ) = 𝐿(𝑥1 ) − 𝐿(𝑥2 ) = 0, so 𝑥1 − 𝑥2 ∈ 𝐾𝑒𝑟(𝐿), but 𝑥1 − 𝑥2 ≠ 0𝑉 .
This is a contradiction, Q.E.D.
a.) Each one-to one mapping 𝑓: 𝑈 → 𝑉 has its inverse map 𝑓 −1 : 𝐼𝑚(𝑓) → 𝑈 such that
𝑓 −1 ∘ 𝑓 = 𝐼𝑑. We need to prove that 1) 𝐼𝑚(𝐿) = ℝ𝑛 and 2) 𝐿−1 is linear.
1) is true because the image of a linear one-to-one transformation has the same dimension
as the initial space (i.e. 𝑛). And there is only one subspace of ℝ𝑛 with dimension 𝑛 (it is
entire ℝ𝑛 itself).
2) Prove that 𝐿−1 (𝑎1 𝑦1 + 𝑎2 𝑦2 ) = 𝑎1 𝐿−1 (𝑦1 ) + 𝑎2 𝐿−1 (𝑦2 ).
Indeed, apply 𝐿 to both sides and get 𝐿[𝐿−1 (𝑎1 𝑦1 + 𝑎2 𝑦2 )] = 𝐿[𝑎1 𝐿−1 (𝑦1 ) + 𝑎2 𝐿−1 (𝑦2 )],
which is true because both parts are equal to 𝑎1 𝑦1 + 𝑎2 𝑦2 .
Because 𝐿 is one-to-one, this means the arguments of 𝐿 are the same, Q.E.D.
b.), c.) Denote matrix for 𝐿 as 𝐴 and matrix for 𝐿−1 as 𝐵 and prove that
matrix for 𝐿−1 𝐿 is 𝐵𝐴. This will prove that 𝐵𝐴 = 𝐼, so 𝐴 is invertible
and that the matrix for 𝐿−1 is 𝐴−1 .
In other words, we need to prove that mu...