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
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asked 2020-11-05
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
\(det(T^{-1}) =\ \frac{1}{det(T)}.\)

Answers (1)

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)
\((det(T^{-1}) = det(S) = \frac{1}{det(T)=det(T)^{-1}} as\ required.\)

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