# 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{

Let be a linear transformatiom and let
$\mathcal{B}$ be a basis of U. Define the determanant det (T) of T as

Show that det (T) is well=defined, i.e. that it does not depend on the choice of the basis $\mathcal{B}.$
Prove that T is invertible if and only if det If T is invertible, show that
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To establish some basic facts about the determinant of a linear transformation T from U to U (U a vector space)
A linear transformation is a map satisfying the properties shown.
(U a vector space) is linear if
and
$T\left(cv\right)=cT\left(v\right),$ 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 be a in basis for U.
(The vectors in B are linearly independent and they span U).
Then any linear map can be represented as a (unique) square matrix:
difined by
$T\left({v}_{i}\right)=\sum _{j=1}^{n}{b}_{ji}{v}_{j}$
With the above notation, we define det (T) = determinant of the matrix
${T}_{b}=\left[{b}_{ij}\right]$
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
$Det\left({T}_{B}\right)=det\left(\left[{b}_{ij}\right]\right)=det\left(\left[{c}_{ij}\right]\right)=Det\left({T}_{C}\right)$
Let B, C are rwo bases of U.
We need to show $det\left({T}_{B}\right)=det\left({T}_{C}\right),$
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
$det\left({T}_{B}\right)=det\left({P}^{-1}{T}_{c}P\right)=det\left(P{\right)}^{-1}det\left({T}_{B}\right)det\left(P\right)=det\left({T}_{C}\right)$
Hence, the proof
Proof of T is invertible iff det (T) is not zero

Now, $S={T}^{-1}$ and
$det\left(T\right)det\left(S\right)=1$ (from the above)