Let B = {V1,..., vn} be an ordered basis for R<sup>n</sup>. Let V<sub>0</sub> = {0}, and for 1 ≤ k ≤ n let
V<sub>k</sub> = span{v<sub>1</sub>,..., v<sub>k</sub>}.
You may assume that V<sub>k</sub> is a subspace of R<sup>n</sup> of dimension k, for all 0 ≤ k ≤ n.
Let A be an n × n matrix, and for all 0 ≤ k ≤ n, define A(V<sub>k</sub>) = {Av | v ∈ V<sub>k</sub>}.
(a) Prove that:
(i) for all v, w ∈ R<sup>n</sup>, [v + w]<sub>B</sub> = [v]<sub>B</sub> + [w]<sub>B</sub>.
(ii) for all v ∈ R<sup>n</sup> and all c ∈ R, [cv]<sub>B</sub> = c[v]<sub>B</sub>.
(b) Prove that for all 1 ≤ k ≤ n, A(V<sub>k</sub>) is a subspace of R<sup>n</sup>.
(c) Suppose A(V<sub>k</sub>) = V<sub>k</sub> for all 1 ≤ k ≤ n. Prove that for all 1 ≤ k ≤ n,
[AV<sub>k</sub>]<sub>B</sub> =
[
c<sub>1</sub>
:
c<sub>k</sub>
0
:
0
]
where c<sub>1</sub>,..., c<sub>k</sub> ∈ R and c<sub>k</sub> ≠ 0.
(d) Deduce that if A(V<sub>k</sub>) = V<sub>k</sub> for all 1 ≤ k ≤ n, then the matrix A is invertible.
