Let T : U → U be a linear transformatiom and let B be a basis of U. Define the determanant det (T) of T as det(T)=det([T] B). Show that det (T) is well=defined, i.e. that it does not depend on the choice of the basis B. Prove that T is invertible if and only if det (T)≠ 0. If T is invertible, show that det(T−1)= 1det(T).

Question

Let T : U → U be a linear transformatiom and let  B be a basis of U. Define the determanant det (T) of T as  det(T)=det([T] B).  Show that det (T) is well=defined, i.e. that it does not depend on the choice of the basis B.  Prove that T is invertible if and only if det (T)≠ 0. If T is invertible, show that  det(T−1)= 1det(T).

coffentw · Accepted Answer

To establish some basic facts about the determinant of a linear transformation T from U to U (U a vector space) A linear transformation T : U → U is a map satisfying the properties shown. T : U → U (U a vector space) is linear if T(v + w)=T(v) + T(w), and T(cv)=cT(v), for vectors v, w in U and scalars c Let B be a basis of U. Then T is represented as a square matrix of size n (n= dimension of U) Let B={v1, v2, ⋯ vn} be a in basis for U. (The vectors in B are linearly independent and they span U). Then any linear map T : B → can be represented as a (unique) square matrix: TB=[bij](1 ≤ i, j ≤ n), difined by T(vi)=∑j=1nbjivj With the above notation, we define det (T) = determinant of the matrix Tb=[bij] In order that det (T) is well-defined , we need to show that this definition does not depend on the choice of the basis B. In other words, if C is some other basis for U, we need to prove that Det(TB)=det([bij])=det([cij])=Det(TC) Let B, C are rwo bases of U. We need to show det(TB)=det(TC), which in turn proves that det(T) is well - defined. (det(T) is independent of the basis) Here , we have used the standard properties of determinants of matrices. Proof: Let B, C are two bases of U. Then Tb=p−1TCP, where P is the change of basis matrix. Now det(TB)=det(P−1TcP)=det(P)−1det(TB)det(P)=det(TC) Hence, the proof Proof of T is invertible iff det (T) is not zero T : U → invertible ⇔ ∃ S(inverse of T) with TS=I(identity) ⇔ det(TS)=det(I)=1 ⇔ det(T)det(S)=1 ⇔ det(T)≠ 0 Now, S=T−1 and det(T)det(S)=1 (from the above) (det(T−1)=det(S)=1det(T)=det(T)−1as required.

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