Let T : U rightarrow U be a linear transformatiom and let mathscr{B} be a basis of U. Define the determanant det (T) of T as det(T) = det([T] mathscr{

BolkowN 2020-11-05 Answered
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(T1)= 1det(T).
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Expert Answer

coffentw
Answered 2020-11-06 Author has 103 answers

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=p1TCP,
where P is the change of basis matrix.
Now
det(TB)=det(P1TcP)=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=T1 and
det(T)det(S)=1 (from the above)
(det(T1)=det(S)=1det(T)=det(T)1as required.

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