Show that det (T) is well=defined, i.e. that it does not depend on the choice of the basis
Prove that T is invertible if and only if det
To establish some basic facts about the determinant of a linear transformation T from U to U (U a vector space)
A linear transformation
Let B be a basis of U. Then T is represented as a square matrix of size n (n= dimension of U)
(The vectors in B are linearly independent and they span U).
Then any linear map
With the above notation, we define det (T) = determinant of the matrix
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
Let B, C are rwo bases of U.
We need to show
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.
where P is the change of basis matrix.
Hence, the proof
Proof of T is invertible iff det (T) is not zero