Question # 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{

Transformation properties
ANSWERED 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
$$det(T) = det([T]\ \mathscr{B}).$$
Show that det (T) is well=defined, i.e. that it does not depend on the choice of the basis $$\mathscr{B}.$$
Prove that T is invertible if and only if det $$(T) \neq\ 0.$$ If T is invertible, show that
$$det(T^{-1}) =\ \frac{1}{det(T)}.$$ 2020-11-06

To establish some basic facts about the determinant of a linear transformation T from U to U (U a vector space)
A linear transformation $$T\ :\ U\ \rightarrow\ U$$ is a map satisfying the properties shown.
$$T\ :\ U\ \rightarrow\ U$$ (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 $$B = \{v_{1},\ v_{2},\ \cdots\ v_{n}\}$$ be a in basis for U.
(The vectors in B are linearly independent and they span U).
Then any linear map $$T\ :\ B\ \rightarrow$$ can be represented as a (unique) square matrix:
$$T_{B}=[b_{ij}](1\ \leq\ i,\ j\ \leq\ n),$$ difined by
$$T(v_{i})=\sum_{j=1}^{n}b_{ji}v_{j}$$
With the above notation, we define det (T) = determinant of the matrix
$$T_{b} = [b_{ij}]$$
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
$$Det (T_{B}) = det([b_{ij}]) = det([c_{ij}]) = Det (T_{C})$$
Let B, C are rwo bases of U.
We need to show $$det(T_{B}) = det(T_{C}),$$
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 $$T_{b} = p^{-1} T_{C} P,$$
where P is the change of basis matrix.
Now
$$det(T_{B}) = det(P^{-1} T_{c} P) = det(P)^{-1} det(T_{B}) det(P) = det(T_{C})$$
Hence, the proof
Proof of T is invertible iff det (T) is not zero
$$T\ :\ U\ \rightarrow\ invertible$$
$$\Leftrightarrow\ \exists\ S(inverse\ of\ T)\ with\ TS=I(identity)$$
$$\Leftrightarrow\ det(TS) = det(I) = 1$$
$$\Leftrightarrow\ det (T) det (S) = 1$$
$$\Leftrightarrow\ det (T) \neq\ 0$$
Now, $$S = T^{-1}$$ and
$$det(T) det(S) = 1$$ (from the above)
$$(det(T^{-1}) = det(S) = \frac{1}{det(T)=det(T)^{-1}} as\ required.$$