We proved a theorem stating that for a linear mapping α : U → V , we havedim Kerα + dim Imα = dimU.A part of the proof went as follows. Assuming dim Kerα = m, there is a basis of Kerα consisting ofm vectors; denote them by u1, . . . , um. Since dim Imα = n, there is a basis of Imα consisting of nvectors; denote them by v1, . . . , vn. Since v1 is in Imα, there is w1 ∈ U such that α(w1) = v1, andsimilarly for all other vectors in the set v1, . . . , vn; in this way, we define n vectors w1, . . . , wn.Now consider a vector u ∈ U . Since α(u) is in Imα, we have α(u) = d1v1 + · · · + dnvn for somenumbers d1, . . . , dn. Consider a vector u′ ∈ U defined as u′ = d1w1 + · · · + dnwn.Prove that u − u′ ∈ Kerα.

rove that the image and the kernel of a linear mapping α : V → W are subspaces of V or W . Inyour answer, you should state clearly whether the image or the kernel is a subspace of V or W . Inyour proof, you may use the definitions of the image and the kernel and the standard properties of asubspace. You may refer to the fact that each linear mapping maps the zero vector to the zero vector(which you proved in question 7).

Let V be a vector space of dimension n and let A = {v1, ..., vn} be an ordered basisof V . Suppose w1, ..., wn ∈ V and let (a1j , ..., anj )t denote the coordinates of wjwith respect to A. Put C = [aij ]. Then show that w1, ..., wn is a basis of V if andonly if C is invertible

This problem provides a useful way to construct invertible linear maps between subspaces X ⊂ Rn and Y ⊂ Rm.3 Let B = {u1, u2, · · · , uk} be a basis of X. (a) Let {v1, v2, · · · , vk} ⊂ Y . Show that there exists a unique linear transform T : X → Y such that T(uj) = vj for every 1 ≤ j ≤ k. To be specific, you need to 1) define a function4 T : X → Y and verify that it is linear, 2) show that if T0 : X → Y is a linear map such that T0(uj) = vj, then T = T0.

Let 𝑇: 𝑈 ->𝑉 be a linear transformation, then rank 𝑇 + 𝑛𝑢𝑙𝑙𝑖𝑡𝑦 𝑇 =

Let T be a linear transformation from an n dimensional vector space V to an mdimensional vector space W and let C be the matrix of T with respect to a basis Aof V and B of W . Show that(a) rank(T ) = rank(C);