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

Ramsey

Ramsey

Answered question

2021-02-11

Let T:UU 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 (T1)=1det(T)

Answer & Explanation

wornoutwomanC

wornoutwomanC

Skilled2021-02-12Added 81 answers

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,UU 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]1i,jn, 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=P1TCP,
where P is the change of basis matrix.
Now
det (TB)=det(P1TCP)=det(P)1det(P)=det(TC)
Step 6
Proof of T is invertible iff det (T) is not zero
T:UU 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=T1 and
det (T)det(S)=1 (from the above)
det (T1)=det(S)=1det(T)=det(T)1,
as required.

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