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

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

wornoutwomanC · Accepted Answer

Step 1 To establish some basic facts about the determinant of a linear transformation T from U to U (U a vector space). Step 2 A linear transformation T,U→U is a map satisfying the properties shown. T: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 Step 3 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 anu linear map T : B → B can be represented as a (unique) square matrix Tb=[bij]1≤i,j≤n, defined by T(v1)=∑j=1nbjivj Step 4 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 two bases of U. We need to show det (TB)=det(TC), which in turn proves that (T) is well - defined. (det (T) is independent of the basis) Step 5 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(P)=det(TC) Step 6 Proof of T is invertible iff det (T) is not zero T:U→U invertible ⇔EES(inverse of T) with TS=I (identity) ⇔ det (TS)=det(I)=1 ⇔det(T)det(S)=1 ⇔det(T)≠0 Step 7 Now, S=T−1 and det (T)det(S)=1 (from the above) det (T−1)=det(S)=1det(T)=det(T)−1, as required.

Let T : U rightarrow U be a linear transformation and let beta be a basis of U Define the determinant det(T) of T as det(T) = det([T]_{beta}). Show ta

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