Question

# 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

Transformation properties
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 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 $$(T) \neq 0.$$ If T is invertible, show that det $$(T^{-1}) = \frac{1}{det(T)}$$

2021-02-12

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 $$T, U \rightarrow U$$ is a map satisfying the properties shown.
$$T : U \rightarrow$$ (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 $$B = {v_{1}, v_{2}, ..., v_{n}}$$ be a in basis for U.
(The vectors in B are linearly independent and they span U).
Then anu linear map T : B \rightarrow B can be represented as a (unique) square matrix $$T_{b} = [b_{ij}]{1 \leq i, j \leq n},$$ defined by
$$T (v_{1}) = \sum_{j = 1}^{n} b_{ji} v_{j}$$
Step 4
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 two bases of U.
We need to show det $$(T_{B}) = det (T_{C}),$$
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 $$T_{B} = P^{-1} T_{C} P,$$
where P is the change of basis matrix.
Now
det $$(T_{B}) = det(P^{-1} T_{C} P) = det (P)^{-1} det (P) = det (T_{C})$$
Step 6
Proof of T is invertible iff det (T) is not zero
$$T : U \rightarrow U$$ invertible
$$\Leftrightarrow EE S (inverse\ of\ T)\ with\ TS = I$$ (identity)
$$\Leftrightarrow\ det\ (TS) = det (I) = 1$$
$$\Leftrightarrow det (T) det (S) = 1$$
$$\Leftrightarrow det (T) \neq 0$$
Step 7
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.