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);

et T and S be linear transformations on a finite dimensional vector space V, where S is non-singular. Then,ans.Rank (ST)= Rank(TS) ≠ Rank TRank (ST) ≠ Rank(TS) ≠ Rank TRank (ST) ≠ Rank(TS) = Rank TRank (ST) = Rank(TS) = Rank T Previous Marked for Review Next

The determinant rank of an 𝑚×𝑛 matrix A is equal to the ___ of A

We proved a theorem stating that for a linear mapping α : U → V , we havedim Kerα + dim Imα = dimU

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.